Two-Engel Lie identity forces class two away from characteristic three ($[x,[x,y]]=0⟹[L,[L,L]]=0(\mathrm{char}K\ne 3)$)

Polarizing the displayed Lie algebra identity gives $[x,[y,z]]+[y,[x,z]]=0$. Antisymmetry shows that the three cyclic terms in the Jacobi identity are all equal. Their sum is zero, so three times each term is zero. Away from characteristic three they therefore vanish, giving nilpotency class at most two. No division by two occurs, so this argument also applies in characteristic two for alternating Lie brackets.