It is provable, using transfer theory and Thompson's Jsubgroup, that if a nonabelian finite simple group $G$ has a nilpotent maximal subgroup $M$, that $M$ must be a 2subgroup. It then follows from Sylow theory that $M$ must be a Sylow 2subgroup of $G$. (It also follows from Sylow theory that, if this is true, any Sylow 2subgroup of $G$ will be maximal.)
It's not unusual, to my understanding, for group theory textbooks to present $PSL_{2}(\mathbb{F}_{17})$ as an example of a nonabelian finite simple group in which the Sylow 2subgroups are maximal. But this can be generalized: $PSL_{2}(\mathbb{F}_{p})$ has maximal Sylow 2subgroups whenever $p$ is prime, $p \geq 17$, and $p$ is 1 away from a power of 2 (which is to say that $p$ is a Fermat prime or a Mersenne prime). (Note that the only composite primepower that is 1 away from a power of 2 is 9. 9 does not need to be included here because $PSL_{2}(\mathbb{F}_{9}) \cong A_{6}$ does not have maximal Sylow 2subgroups. So weakening the condition by allowing powers of primes does not help here.)
My question is: if a nonabelian finite simple group has maximal Sylow 2subgroups, it is necessarily isomorphic to one of the $PSL_{2}$ groups described above?

1$\begingroup$ I think the answer can be deduced from some pretty heavy duty theorems (eg M. Aschbacher's $C(G,T)$theorem). There may be more direct proofs available, $\endgroup$– Geoff RobinsonJun 16, 2017 at 8:28
1 Answer
Yes this follows from work of Baumann and Thompson. See the paper `On finite insoluble groups with nilpotent maximal subgroups' by John Rose https://doi.org/10.1016/00218693(77)903015
In fact, these are the only examples of simple groups with nilpotent maximal subgroups.