The maximal subgroups of $A_{7}$ are $A_{6}$ (index 7), $S_{5}$ (index 21), $(A_{4} \times A_{3}):2$ (index 35), and $GL_{3}(2)$ (index 15).

If $G$ is a nonabelian finite simple group and all maximal subgroups of $G$ have odd index, does it follow that $G \cong A_{7}$?

I don't know the Lie type groups well enough to answer this for myself, but here I'll prove it for the alternating groups (and a case-by-case check works for the sporadic groups):

Assume that $n \geq 5$. If $n+1$ is not a power of 2, then choose an $r < \frac{n}{2}$ such that $\binom{n}{r}$ is even; then the stabilizer of an $r$-element subset of $\{ 1, \ldots, n \}$ in $A_{n}$ is a maximal subgroup of even index. Otherwise, write $n = 2^{k}-1$. Since we are assuming $n \neq 7$, we say that $k \geq 4$. Then $GL_{k}(2)$, in its action on vectors of $\mathbb{F}_{2}^{k} \setminus \{ 0 \}$, is a subgroup of $A_{n}$. Since $GL_{k}(2)$ is a proper subgroup of $A_{n}$, consider a maximal subgroup $M$ containing $GL_{k}(2)$.

Claim. $[A_{n}:M]$ is even.

Proof of Claim. $GL_{k}(2)$ acts doubly transitively on the indices, so $M$ does too. Therefore $M$ acts primitively on the indices. If $[A_{n}:M]$ were odd, then every 2-element of $A_{n}$ would be conjugate to an element of $M$; in particular, $M$ would contain permutations of cycle shape $(2,2)$. But $n = 2^{k}-1 \geq 15$ and, on more than $8$ points, the only primitive permutation groups containing a permutation of cycle shape $(2,2)$ are the symmetric and alternating groups. This is a contradiction because we assumed $M < A_{n}$.