So that this question does not remain listed as unanswered, and for the benefit of any future readers, I will provide an answer.
This follows the suggestions in the comments of Nicky Hekster, to look at the exposition in Derek J.S. Robinson - A Course in the Theory of Groups.
In Robinson this result is
Theorem 12.3.4 (Zorn) A finite Engel group is nilpotent.
To prove it, we need to recall the following: (For the benefit of the reader I shall include the same numbering as in Robinson so that the proofs for these may be looked up if you wish).
Notation and preliminaries: $L(G)$ is the set of left Engel elements of $G$. If $G$ is an Engel group then $G=L(G)$.
The Hirsch-Plotkin radical of $G$ is the unique maximal normal locally nilpotent subgroup, I will denote this as $HP(G)$. The Fitting subgroup is the subgroup generated by all the normal nilpotent subgroups of a group $G$ and is written $Fit(G)$. Note that if $G$ is finite then $Fit(G)$ is nilpotent, and in particular in the case $G$ is finite, if $G=Fit(G)$ we can conclude $G$ is nilpotent.
We also need the following two Theorems:
Theorem 9.1.9 (O.J Schmidt) Assume that every maximal subgroup of a
finite group $G$ is nilpotent but $G$ itself is not nilpotent. Then
$G$ is soluble. (Note that the theorem says more than this but we do
not need the rest).
Theorem 12.3.3 (Gruenberg). Let $G$ be a soluble group. Then $L(G)$ coincides with the Hirsch-Plotkin radical. (Note that the
theorem says more than this but we do not need the rest)
We can now prove the result.
Proof of Theorem 12.3.4
Suppose that this is false. Then let $G$ be a minimal counter example, that is, $G$ is a finite Engel group with smallest order subject to being not nilpotent. Thus every proper subgroup of $G$ must be nilpotent. Thus $G$ is soluble by the Theorem of Schmidt. Then by the Theorem of Gruenberg above, we know that if $G$ is soluble then $L(G)$ coincides with the Hirsch-Plotkin radical. Since $G$ is finite, the Hirsch-Plotkin radical of $G$ coincides with the Fitting subgroup of $G$, and since $G$ is Engel we know that $G=L(G)$. Thus we have $$G=L(G)=HP(G)=Fit(G).$$ Thus $G$ must be nilpotent. $\square$