= Solution
Polarizing $[x,[x,z]]=0$ gives
$$
[x,[y,z]]+[y,[x,z]]=0.
$$
Let $T=[x,[y,z]]$. Antisymmetry and this identity give $[y,[z,x]]=T$ and $[z,[x,y]]=T$. The <Jacobi identity> therefore gives $3T=0$. Since the characteristic is not three, $T=0$ for every triple. Consequently
$$
\boxed{[L,[L,L]]=0,\qquad L\text{ has nilpotency class at most two}.}
$$
No division by two was used, so the proof also applies in characteristic two with the usual alternating definition of a <Lie bracket>.
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