Find all positive integer $n$ such that there exist $m\in\mathbb{Z}$ with $2^n-1|m^2+9$ I have an problem with elementary number theory:
Find all positive integer $n$ such that there exist $m\in\mathbb{Z}$ with $2^n-1|m^2+9$
It's look like the problem in this link, but there some differences: Find all positive integer $n$ such that there exists $m$ with $2^n-1|m^2+17^2$.

If $2^n-1$ has a prime divisor $p\equiv3\pmod4$, then $m^2+3^2\equiv0\pmod p$. If $p\neq 3$, let $a$ be an inverse of $3$ modulo $p$. Then $(am)^2\equiv-1\pmod p$, contradicting $p\equiv3\pmod4$.
Thus $3$ must be the only prime divisor of $2^n-1$ that is $3$ mod $4$. First, $n=1$ clearly works. Suppose $n \geq 2$. Then $2^n-1$ has at least one prime divisor congruent to $3$ mod $4$, which must be $3$. Thus $n$ is even. Write $n = 2k$. Then $2^k-1 \mid m^2+9$, so again, $k=1$ or $k$ is even... By induction, we see that $n$ must be a power of $2$. Write $n=2^m$, then
$$2^{2^m}-1 = 3 \cdot \prod_{k=1}^{m-1}\left( 2^{2^k}+1 \right)$$
Because every prime divisor of the second factor is $1$ mod $4$, each $n=2^m$ is a solution.
