We first show that if $2^n - 1$ is a divisor of $m^2 + 9$ for some number $m$, then $n$ is the power of number 2. Otherwise the $n$ has some odd divisor $\ell \ge 3$ and so $2^\ell - 1$ is a divisor of $m^2 + 9$ because $2^\ell - 1$ divide $2^n - 1$. But for $\ell \ge 3$ we have $2^\ell - 1 \equiv -1 \pmod 4$, and therefore the $2^\ell - 1$ has prime divisors $p$ such that the $p \equiv -1 \pmod 4$. Firstly, we deal with $p \ne 3$. Since $p \mid m^2 + 3^2$ by Fermat little theorem we have
$$1 \equiv m^{p-1} \equiv (m^2)^{(p-1)/2} \equiv (-9)^{(p-1)/2} \equiv (-1)^{(p-1)/2} \cdot 3^{p-1} \equiv -1 \pmod{p}$$
But this is impossible, so $p=3$. But then for odd $\ell$ we have $2^\ell \equiv -1 \not\equiv -1 \pmod 3$, which is a contradiction. Consider $n=2^k$ now. For $n=1$, the claim is true. If $k \ge 1$ then
$$2^n - 1 = 3(2^2 + 1)(2^{2^2} + 1)\cdots(2^{2^{k-1}} + 1)$$
Therefore, if $2^n - 1$ divides $m^2 + 9$, also $2^\ell + 1$ divides $m^2 + 9$ for each $\ell = 1, 2, \dots, k-1$. Moreover if $\alpha \ne \beta$ the numbers $2^{2\alpha} + 1$ and $2^{2\beta} + 1$ are relatively prime because if $d > 2$ is the greatest common divisor and $\alpha > \beta$, then
$$-1 \equiv 2^{2\alpha} \equiv (2^{2\beta})^{2^{\alpha-\beta}} \equiv 1 \pmod{d}$$
which is a contradiction. So it is necessarily $d=1$ ($d \ne 2$, because both are odd).
Then, according to the Chinese remainder theorem there is a natural number $c$ such that
$$c \equiv 2^{2\ell} \pmod{2^{2^{\ell+1}}}, \quad \forall \ell = 0, 1, \dots, k-2$$
Then $c^2 + 1 \equiv 0 \pmod{2^{2\ell+1}}$ for $\ell = 0, 1, \dots, k-2$, and thus $2^n - 1$ divides $(3c)^2 + 9$. $\square$

We claim that, if $2^n - 1$ divides $m^2 + 9$ for some $m \in \mathbb{N}$, then $n$ must be a power of 2. Suppose otherwise that $n$ has an odd divisor $d > 1$. Then $2^d - 1 \mid 2^n - 1$ is also a divisor of $m^2 + 9 = m^2 + 3^2$. However, $2^d - 1$ has some prime divisor $p$ of the form $4k-1$, and by a well-known $p$ divides both $m$ and 3. Hence $p=3$ divides $2^d - 1$, which is impossible because, for $d$ odd, $2^d \equiv 2 \pmod 3$. Hence $n = 2^r$ for some $r \in \mathbb{N}$.
Now let $n = 2^r$. We prove the existence of $m$ by induction on $r$. The case $r=1$ is trivial. Now for any $r > 1$ note that
$$2^{2r} - 1 = (2^{2^{r-1}} - 1)(2^{2^{r-1}} + 1)$$
The induction hypothesis claims that there exists an $m_1$ such that $2^{2^{r-1}} - 1 \mid m_1^2 + 9$. We also observe that $2^{2^{r-1}} + 1 \mid m_2^2 + 9$ for simple $m_2 = 3 \cdot 2^{2^{r-2}}$. By the Chinese remainder theorem, there is an $m \in \mathbb{N}$ that satisfies $m \equiv m_1 \pmod{2^{2^{r-1}} - 1}$ and $m \equiv m_2 \pmod{2^{2^{r-1}} + 1}$. It is easy to see that this $m^2 + 9$ will be divisible by both $2^{2^{r-1}} - 1$ and $2^{2^{r-1}} + 1$.
This completes the induction. $\square$