For which finite groups $M$ is it the case that every finite group $G$ with $M$ as a maximal subgroup solvable?
If $M$ satisfies this condition then $M$ is solvable. Also, if $M$ is abelian then $M$ satisfies this condition. Futhermore, I believe that if $M$ is nilpotent and if all 2-subgroups of $M$ are normal subgroups of $M$ (if Sylow 2-subgroups of $M$ are abelian or quaternion, for example) then $M$ satisfies this condition (proof below).
More specific questions:
1) Is there a non-nilpotent group that satisfies this condition?
2) Which 2-groups satisfy this condition?
Apparently, the dihedral group of order 8 satisfies this condition (see Mikko Korhonen's comment on this post). Also, if $M\times N$ satisfies this condition then $M$ and $N$ both satisfy this condition.
(This proof is adapted from j.p.'s answer to the linked question). Let $G$ be minimal such that $G$ is not solvable and such that $G$ contains a maximal subgroup $M$ that is nilpotent and whose 2-subgroups are normal. If $M$ contains a nontrivial normal subgroup $N$ of $G$ then $G/N$ contradicts the minimality of $G$. Thus, $M$ does not contain nontrivial normal subgroups of $G$. In particular, $N_G(P)=M$ for all Sylow $p$-subgroups $P$ of $M$. Then $P$ is a Sylow $p$-subgroup of $N_G(P)$ so $P$ is a Sylow $p$-subgroup of $G$. This shows that $M$ is a Hall subgroup of $G$.
If $P$ is a Sylow $p$-subgroup of $M$ and if $Q$ is a nontrivial normal subgroup of $P$ then $N_G(Q)=M$ which has a normal $p$-complement. For $p=2$, Frobenius' normal $p$-complement theorem gives that $G$ has a normal $p$-complement. For $p\geq3$, Thompson's normal $p$-complement theorem or Glauberman's normal $p$-complement theorem gives that $G$ has a normal $p$-complement (since you only have to consider characteristic $p$-subgroups).
Thus, for each prime $p$ dividing the order of $M$, $G$ has a normal $p$-complement. Then $M$ has a normal complement $N$ in $G$. Since $M$ is solvable but $G$ is not solvable, $N$ is not solvable. In particular, $N$ does not admit a fixed-point-free automorphism of prime order. If $m\in Z(M)$ has prime order then $C_N(m)$ is nontrivial. Then $C_N(m)M$ is a subgroup of $G$ that properly contains $M$ so $C_N(m)M=G$ by the maximality of $M$. Comparing cardinalities shows that $C_N(m)=N$ so $m\in Z(G)$. Then $\langle m\rangle$ is a nontrivial normal subgroup of $G$ contained in $M$ which is a contradiction.
