$a = b$ is the only possibility.

First, substituting $n = 2$, we must have $1 + 2^a \mid 1 + 2^b$. Let $b = aq + r$, where $0 \le r < a$, then we know
$$1 + 2^b = 1 + 2^{aq + r} \equiv 1 + (-1)^q \times 2^r \pmod{1 + 2^a}.$$
Since $|1 + (-1)^q \times 2^r| < 1 + 2^a$, we must have $1 + (-1)^q \times 2^r = 0$, so $q$ is odd and $r = 0$. In other words, $b$ must be an odd multiple of $a$.

We now prove that $q = 1$. Suppose for contradiction that $q$ has an odd prime factor $p$, and let $q = lp$, so $b = lpa$. Take $n = 2^{p-1}$, then we must have
$$\frac{2^{pa} - 1}{2^a - 1} = 1 + 2^a + \cdots + 2^{(p-1)a} = f_a(n) \mid f_b(n) = 1 + 2^b + \cdots + 2^{(p-1)b}. \quad (1)$$
However,
$$1 + 2^b + \cdots + 2^{(p-1)b} \equiv 1 + 1 + \cdots + 1 = p \pmod{2^{pa} - 1},$$
and $0 < p < 1 + 2^a + \cdots + 2^{(p-1)a}$, so the divisibility in (1) is impossible, a contradiction.

Therefore $q$ cannot have an odd prime factor, that is, $q = 1$. This completes the proof.