Induction.
The base case $n = 1$ reads
$2^3 \mid a^2 - 1; \tag 1$
this holds since, $a$ being odd, $a - 1$ and $a + 1$ are both even; since $a \pm 1$ are consecutive even integers, each is divisible by $2$ and one is divisible by $4 = 2^2$; this establishes (1).
We hypothesize
$2^{k + 2} \mid a^{2^k} - 1; \tag 2$
we have
$a^{2^{k + 1}} - 1 = (a^{2^k} - 1)(a^{2^k} + 1); \tag 3$
we also have
$2 \mid a^{2^k} + 1, \tag 4$
since $a$ is odd, implying $a^{2^k}$ is also odd; (2) and (4) imply
$\exists c, d \in \Bbb Z, \; 2^{k + 2}c = a^{2^k} - 1, \; 2d = a^{2^k} + 1; \tag 5$
thus, via (3),
$2^{(k + 1) + 2}cd = 2^{(k + 2) + 1}cd =(2^{k + 2}c)(2d) = (a^{2^k} - 1)(a^{2^k} + 1) = a^{2^{k + 1}} - 1; \tag 6$
and thus,
$2^{(k + 1) + 2} \mid a^{2^{k + 1}} - 1, \tag 7$
completing the induction and proving that
$2^{n + 2} \mid a^{2^n} - 1, \; \forall n \in \Bbb N. \tag 8$