= Solution
For the <Adjoint representation of a Lie algebra>, the operator associated with $x\in L$ is
$$
\operatorname{ad}_x(y)=[x,y].
$$
The <Jacobi identity> gives $[\operatorname{ad}_x,\operatorname{ad}_y]=\operatorname{ad}_{[x,y]}$, so this really is a <Lie algebra representation>. Define the <lower central series of a Lie algebra> by $L^1=L$ and $L^{r+1}=[L,L^r]$. A <nilpotent Lie algebra> is one for which $L^{c+1}=0$ for some integer $c$. If this holds, $\operatorname{ad}_x$ carries $L^r$ into $L^{r+1}$, and consequently $(\operatorname{ad}_x)^cL=0$. This proves one implication without any condition on the field.
For the converse, we prove the linear form of the <Engel theorem>: if $A\subseteq\operatorname{End}(V)$ is a finite-dimensional <Lie algebra> consisting entirely of <nilpotent linear maps>, with $V\ne0$ finite-dimensional, then there is a nonzero vector annihilated by all of $A$. This argument works over any field.
Induct on $\dim A$. The assertion is immediate when $A=0$. First observe that if $b^m=0$, then its <commutator> action on $\operatorname{End}(V)$ is nilpotent, because
$$
(\operatorname{ad}_b)^r(T)=\sum_{j=0}^r(-1)^j\binom rj b^{r-j}Tb^j=0\qquad(r\ge2m-1).
$$
Let $B$ be a proper <Lie subalgebra> of $A$. The <commutator> action of $B$ on $A/B$ therefore consists of <nilpotent linear maps>. Its image has dimension at most $\dim B<\dim A$, so induction supplies a nonzero coset $a+B$ annihilated by $B$. Thus $[B,a]\subseteq B$, and the normalizer $N_A(B)=\{a\in A:[a,B]\subseteq B\}$ strictly contains $B$. This is the <Engel normalizer lemma>.
Choose $B$ maximal among proper <Lie subalgebras>. Its normalizer must be $A$, so $B$ is a <Lie algebra ideal>. Moreover $\dim(A/B)=1$: otherwise the inverse image of a one-dimensional subalgebra of $A/B$ would lie strictly between $B$ and $A$. By induction,
$$
W=\{v\in V:bv=0\text{ for all }b\in B\}\ne0.
$$
Since $B$ is an ideal, $W$ is $A$-invariant: $b(av)=a(bv)+[b,a]v=0$. Take $x\in A\setminus B$. Its restriction to $W$ is nilpotent and hence has a nonzero kernel. A nonzero vector in that kernel is annihilated by $B$ and $x$, and therefore by all of $A$. This completes the induction.
Apply this result successively to $V$, then its quotient by a common annihilated line, and so on. We obtain a complete flag $0=V_0\subset V_1\subset\cdots\subset V_n=V$ with $AV_i\subseteq V_{i-1}$. Equivalently every element of $A$ is strictly upper triangular in one common basis; any product of $n$ such operators is zero.
Now take $V=L$ and $A=\operatorname{ad}(L)$. The hypothesis supplies exactly the nilpotence needed above. Every iterated bracket of length $n+1$, where $n=\dim L$, is an application of $n$ adjoint operators and vanishes. These brackets span $L^{n+1}$, giving $L^{n+1}=0$. The zero algebra is already nilpotent. Thus \b[$L$ is nilpotent if and only if every $\operatorname{ad}_x$ is nilpotent]. The common flag is essential: separate nilpotence of unrelated operators would not imply that their mixed products vanish.
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