= Solution
In the finite-dimensional setting, let $R(a,b)$ denote the irreducible $\mathfrak{sl}_3$ <highest-weight representation> of highest weight $aL_1-bL_3$, where $a,b$ are nonnegative <integers>. Thus $(a,b)$ are its <Dynkin labels>. The target $\lambda=L_1-2L_3$ has labels $(1,2)$. Write the highest-weight difference in <simple roots> as
$$
(aL_1-bL_3)-\lambda=r(L_1-L_2)+s(L_2-L_3),\qquad
r=\frac{2a+b-4}{3},\quad s=\frac{a+2b-5}{3}.
$$
The <dominant weight multiplicity formula for sl3> gives zero unless $r,s$ are nonnegative <integers>, and otherwise gives
$$
\dim R(a,b)_\lambda=1+\min(a,b,r,s).
$$
For completeness, this multiplicity is an actual count. In the <sl3 interlacing character formula>, use top row $(a+b,b,0)$ and bottom entry $s+3$; the target diagonal exponents are $(s+3,s+2,s)$. The middle row $(P,Q)$ obeys $P+Q=2s+5$ and
$$
\max\{b,2s+5-b,s+3\}\leq P\leq\min\{a+b,2s+5\}.
$$
Each allowable integral $P$ gives one <Gelfand–Tsetlin basis> vector. Subtracting the lower bound from the upper bound gives the minimum of $a,s,r,r+3,b,s+2$, which reduces to $\min(a,b,r,s)$. This proves the displayed multiplicity formula in the case at hand.
Since both summands have a nonzero <weight space> and their dimensions at $\lambda$ sum to three, their multiplicities must be one and two. For multiplicity one, the minimum is zero. The cases $a=0$ and $b=0$ give $(0,4+3j)$ and $(5+3j,0)$; the remaining cases $r=0$ or $s=0$ give $(1,2)$ and $(3,1)$. Hence the full family is
$$
\mathcal C_1=\{R(0,4+3j),R(5+3j,0):j\geq0\}\cup\{R(1,2),R(3,1)\}.
$$
For multiplicity two, the minimum is one. The cases $a=1$ and $b=1$ give $(1,5+3j)$ and $(6+3j,1)$; the remaining cases $r=1$ or $s=1$ give $(2,3)$ and $(4,2)$. Thus
$$
\mathcal C_2=\{R(1,5+3j),R(6+3j,1):j\geq0\}\cup\{R(2,3),R(4,2)\}.
$$
The minimum conditions exhaust all cases and each listed module has the stated multiplicity. \b[All candidate unordered pairs are $\boxed{\{U,V\}\text{ with }U\in\mathcal C_1,\ V\in\mathcal C_2}$.] There are infinitely many at this stage; the question imposes no upper bound on their highest weights.
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