 Solution

For the Adjoint representation of a Lie algebra, the operator associated with $x\in L$ isThe Jacobi identity gives $[{\mathrm{ad}}_x,{\mathrm{ad}}_y]={\mathrm{ad}}_{[x,y]}$, 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, ${\mathrm{ad}}_x$ carries $L^r$ into $L^{r+1}$, and consequently $({\mathrm{ad}}_x{)}^{c}L=0$. This proves one implication without any condition on the field.

For the converse, we prove the linear form of the Engel theorem: if $A\subseteq \mathrm{End}(V)$ is a finite-dimensional Lie algebra consisting entirely of nilpotent linear maps, with $V\ne 0$ 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 $\mathrm{End}(V)$ is nilpotent, becauseLet $B$ 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,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 $A{V}_i\subseteq V_{i-1}$. 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 $A=\mathrm{ad}(L)$. 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 $L$ is nilpotent if and only if every ${\mathrm{ad}}_x$ is nilpotent. The common flag is essential: separate nilpotence of unrelated operators would not imply that their mixed products vanish.