= Solution
In the finite-dimensional complex <semisimple Lie algebra> setting, the <rank of a semisimple Lie algebra> is the dimension of a <Cartan subalgebra> $\mathfrak h$, equivalently the number of independent commuting Cartan generators. For a compact real algebra, use its complexification and the corresponding maximal torus. Simultaneously diagonalizing the commuting adjoint actions of $\mathfrak h$ gives the <root-space decomposition>
$$
\mathcal L=\mathfrak h\oplus\bigoplus_{\alpha\ne0}\mathcal L_\alpha,\qquad
\mathcal L_\alpha=\{X:[H,X]=\alpha(H)X\text{ for every }H\in\mathfrak h\}.
$$
The nonzero linear functionals $\alpha$ for which this space is nonzero are the <roots of a root system>. An invariant inner product identifies them with vectors in a real Euclidean root space. These are the usual semisimple assumptions behind the rank and root language here.
The displayed generators $H,E^+,E^-$ form an <sl2 triple>. Start with a nonzero lowest-<weight> element $X_0$ and define $X_{n+1}=[E^+,X_n]$. The <Jacobi identity> gives
$$
[H,X_{n+1}]=[[H,E^+],X_n]+[E^+,[H,X_n]].
$$
Induction starting with $[H,X_0]=-\lambda X_0$ consequently gives the <finite sl2 lowest-weight ladder> <weights>
$$
\boxed{[H,X_n]=(2n-\lambda)X_n.}
$$
Now commute the lowering equation with $E^+$ and again use the <Jacobi identity>:
$$
[E^+,[E^-,X_n]]=[[E^+,E^-],X_n]+[E^-,[E^+,X_n]]
=(2n-\lambda)X_n+q_{n+1}X_n.
$$
The left side is $q_nX_n$. Whenever $X_n\ne0$, coefficient comparison gives
$$
\boxed{q_{n+1}=q_n+\lambda-2n.}
$$
The initial condition is $q_0=0$, meaning $[E^-,X_0]=0$; directly $[E^-,X_1]=[-H,X_0]=\lambda X_0$, so $q_1=\lambda$. Summing the recurrence from $0$ to $n-1$ yields
$$
\boxed{q_n=n\lambda-2\sum_{k=0}^{n-1}k=n(\lambda-n+1).}
$$
In fact the vector identity $[E^-,X_n]=n(\lambda-n+1)X_{n-1}$ follows inductively from the same <Jacobi identity> without needing a separately assumed lowering coefficient.
If $X_{n_0}$ is the last nonzero element and $[E^+,X_{n_0}]=0$, apply $E^-$ to this zero next element. The lowering identity gives
$$
0=[E^-,X_{n_0+1}]=(n_0+1)(\lambda-n_0)X_{n_0},
$$
so \b[$\lambda=n_0$], a nonnegative <integer>. The nonzero endpoint is necessary. Literally allowing $X_{n_0}=0$ would make the printed conclusion false: take $X_0=E^-$ in the adjoint <representation> of $\mathfrak{sl}_2$. Then $\lambda=2$, $X_1=H$, $X_2=-2E^+$, $X_3=0$. Choosing $n_0=3$ gives a vanishing next bracket but not $\lambda=3$. The intended endpoint is the first terminating bracket after the nonzero ladder.
For the plus-sign nested <commutators>, use $H=H_1$, $E^\pm=E_1^\pm$ and $X_0=E_2^+$. The cross relation gives $[E_1^-,E_2^+]=0$, while $[H_1,E_2^+]=K_{21}E_2^+$. Thus the lowest-<weight> label is $\lambda=-K_{21}$. If the $n$-fold raised element is nonzero and the $(n+1)$-fold element vanishes, the preceding endpoint argument gives $\lambda=n$ and hence
$$
\boxed{K_{21}=-n.}
$$
For the minus-sign ladder, use the <sl2 triple> $H'=-H_1$, $E'^+=E_1^-$, $E'^-=E_1^+$ and start at $X_0=E_2^-$. Its lowest-<weight> equation is again $[H',X_0]=K_{21}X_0$, and its lowering bracket vanishes by the cross relation. The same argument proves exactly the same result, including the endpoint assumption that is explicit in the nested-<commutator> question.
For arbitrary <roots of a root system> $\alpha,\beta$, choose the normalized <sl2 subalgebra associated with a root> $\beta$, whose generators obey $[H_\beta,E_\beta^\pm]=\pm2E_\beta^\pm$ and $[E_\beta^+,E_\beta^-]=H_\beta$. The <coroot> normalization gives
$$
\alpha(H_\beta)=\frac{2\alpha\cdot\beta}{\beta\cdot\beta}.
$$
The normalization has a concrete construction. Use the nondegenerate <Killing form> $B$, and define $t_\beta\in\mathfrak h$ by $B(t_\beta,H)=\beta(H)$. Invariance makes root spaces orthogonal unless their roots sum to zero, so there exist $e\in\mathcal L_\beta$ and $f\in\mathcal L_{-\beta}$ with $B(e,f)=1$. Invariance then gives $B([e,f],H)=B(e,[f,H])=\beta(H)$, hence $[e,f]=t_\beta$. With $(\beta,\beta)=\beta(t_\beta)$, take $E_\beta^+=e$, $E_\beta^-=2f/(\beta,\beta)$ and $H_\beta=2t_\beta/(\beta,\beta)$. Their brackets have the required normalization and $\alpha(H_\beta)=2(\alpha,\beta)/(\beta,\beta)$. Let $v$ be a nonzero $\alpha$ <root vector> with $H_\beta$ <weight> $w=\alpha(H_\beta)$. Repeatedly apply $\operatorname{ad}E_\beta^-$ until reaching a last nonzero vector $v_0=(\operatorname{ad}E_\beta^-)^pv$. Its <weight> is $w-2p$. This process terminates because different <weights> are linearly independent and the <Lie algebra> is finite-dimensional. Starting at $v_0$, repeatedly raise; this also terminates for the same reason. The proven lowest-<weight> endpoint formula says $w-2p=-N$ for a nonnegative integer $N$. Therefore
$$
\boxed{\frac{2\alpha\cdot\beta}{\beta^2}=w=2p-N\in\mathbb Z.}
$$
This proves integrality of <Cartan integers> directly from the finite ladder, rather than assuming the crystallographic axiom.
Finally use the indices actually printed in the PDF: $E_1^+=R^1{}_2$, $E_2^+=R^2{}_3$, $E_1^-=R^2{}_1$, $E_2^-=R^3{}_2$. The converted TeX has corrupted these indices. The prescribed bracket is that of trace-free <matrix units>. The sum of the diagonal generators is zero, leaving two independent diagonal elements, and their pair brackets give
$$
\boxed{H_1=R^1{}_1-R^2{}_2,\qquad H_2=R^2{}_2-R^3{}_3.}
$$
For example $[R^1{}_2,R^2{}_1]=R^1{}_1-R^2{}_2$, and the analogous second pair gives $H_2$. The cross raising-lowering brackets vanish. A diagonal $H$ with entries $h_r$ has $[H,R^r{}_s]=(h_r-h_s)R^r{}_s$. Thus $H_1$ acts on $E_1^+,E_2^+$ with <weights> $2,-1$, and $H_2$ acts on them with <weights> $-1,2$. The negative generators have the opposite <weights>. In the question's index ordering,
$$
\boxed{(K_{ji})=\begin{pmatrix}2&-1\\-1&2\end{pmatrix}.}
$$
This is the <Cartan matrix> of the <A2 root system>, and the <Lie algebra> generated by these matrix units has rank two. As an additional bracket check, $[E_1^+,E_2^+]=R^1{}_3\ne0$ whereas $[E_1^+,R^1{}_3]=0$, so the nested ladder has $n=1$ and indeed $K_{21}=-1$. The negative ladder likewise terminates after one nonzero bracket.
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