3-Engel and 2-torsion-free implies 2-local class three for Lie rings

Statement
Suppose $$L$$ is a 3-Engel Lie ring, i.e., the following identity holds:

$$[x,[x,[x,y]]] = 0 \ \forall \ x,y \in L$$

Suppose further that $$L$$ has no 2-torsion, i.e., $$2x = 0$$ implies $$x = 0$$ for $$x \in L$$. Then, $$L$$ is a Lie ring of 2-local nilpotency class three, i.e., the 2-local nilpotency class of $$L$$ is at most three.

Facts used

 * 1) uses::Polarization trick: We use the polarization trick in three variables.
 * 2) uses::Higgins' lemma on Engel conditions

Preliminary observations
In order to establish the result, we need to show that all Lie products of length four that involve only two variables must take the value zero. We know that it suffices to restrict attention to right normed expressions because of the Jacobi identity. Up to interchange of $$x$$ and $$y$$ and using skew symmetry in $$x,y$$, we see that there are only two types of expressions: $$[x,[x,[x,y]]]$$ and $$[x,[y,[x,y]]]$$. Thus, it suffices to prove that $$[x,[y,[x,y]]] = 0$$ for all $$x,y \in L$$.

Proof details
Given: A Lie ring $$L$$. We have $$[x,[x,[x,y]]] = 0$$ for all $$x,y \in L$$. We also have that $$L$$ has no 2-torsion: $$2x = 0 \implies x = 0$$ for $$x \in L$$.

To prove: $$[x,[y,[x,y]]] = 0$$ for all $$x,y \in L$$.

Proof: For notational convenience, we will use the string $$abcd$$ to denote the right-normed Lie product $$[a,[b,[c,d]]]$$.

Alternative proof
The result can also be proved using Fact (2) setting $$n= 2$$. That proof is more illuminative because of its potential for generalization.