I make a remark in the opposite direction. Let $p \geq 17$ be a Fermat or Mersenne prime, so that $X = {\rm PSL}(2,p)$ has a dihedral Sylow $2$-subgroup $D$ which is maximal. Let $d >1$ be a power of $2$, and let $Q$ be a transitive $2$-subgroup of $S_{d}$. Let $A= {\rm Aut}(X)$ and let $T$ be a Sylow $2$-subgroup of $A$. Then $T$ is a maximal subgroup of $XT$.
Using this action of $Q$ on $d$ points, form the wreath product $XT \wr Q$. Then $T \wr Q$ is a Sylow $2$-subgroup of $(XT) \wr Q$, and is a maximal subgroup of $(XT) \wr Q$.
This (either $T$ or $T \wr Q$) is close to the general shape of a corefree nilpotent maximal subgroup $G$ of a non-solvable group $H$. Something close to this was proved by Baumann and Rose (although Rose allowed the maximal subgroup not to be corefree, which obscured the corefree case somewhat).
For let $H$ be a non-solvable group with nilpotent and corefree maximal subgroup $G$ (that is to say, $G$ contains no non-trivial normal subgroup of $H$). I won't repeat the argument that $G$ must be a $2$-group, since this has been discussed elsewhere. Then $G$ must be a Sylow $2$-subgroup of $H$, as $G$ is contained in some Sylow $2$-subgroup (which is certainly a proper subgroup of $H$) and since $G$ is maximal.
Now $H = GM$, where $M$ is a non-solvable minimal normal subgroup, so is a direct product of $d$ copies of a non-Abelian simple group $X$. Let $S = G \cap M$, which is a Sylow $2$-subgroup of $M$ (and is non-trivial). Then $S \lhd G$ and we have $H = MN_{H}(S).$ Now we have $G \leq N_{H}(S)$ (and $N_{H}(S)$ is proper), so we must have $G = N_{H}(S).$ In particular $S = G \cap M = N_{M}(S).$
If $d = 1,$ we have (up to isomorphism) $H \leq SX,$ where $S$ is a Sylow $2$-subgroup of ${\rm Aut}(X)$ and $X$ is a simple group with a maximal Sylow $2$-subgroup.
If $d > 1$, there is a normal subgroup $K$ of $H$ which is the kernel of the permutation action of $H$ on its components, and $K$ is a subgroup of the direct product of $d$ copies of $XS_{1}$ containing $X^{d}$, where $S_{1}$ is Sylow $2$-subgroup of ${\rm Aut}_{G}(X).$ Furthermore, $K = X^{d}(G \cap K)$ and $G \cap K$ is a Sylow $2$-subgroup of $K$ which is maximal in $K$.
Now $G \cap K$ is a subgroup direct product of $d$ copies of $S_{1}$, and $G/G \cap K$ is a transitive $2$-subgroup of $S_{d}$.