Suppose that $[L,L]$ is nilpotent. Now $L/[L,L]$ is abelian, hence solvable. It follows that both $[L,L]$ and $L/[L,L]$ are solvable. Then $L$ is solvable, too, because every extension of a solvable Lie algebra by a solvable Lie algebra is itself solvable (we have the extension $0\rightarrow [L,L]\rightarrow L\rightarrow L/[L,L]\rightarrow 0$). The argument is also valid over fields of characteristic $p>0$, whereas the converse statement, i.e., that $L$ solvable implies $[L,L]$ nilpotent, need not be true in that case.