3. Find all positive integers $n$, such that for all $n$, there exists $m$ such that $2^{n}-1 \mid m^{2}+9$.

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3. Notice that the prime factors of $m^{2}+9$, except for 3, must be primes of the form $4k+1$. This is because if $p$ is a prime factor of $m^{2}+9$, then $\left(\frac{-9}{p}\right)=1$, or $p=3$.
Let $n=2^{s} \cdot t$, where $t$ is an odd number. If $t>1$, then $2^{t}-1\left|2^{n}-1\right| m^{2}+9$.

Notice that $2^{t}-1 \equiv(-1)^{t}-1 \equiv-2(\bmod 3)$, meaning $2^{t}-1$ is not a multiple of 3, so the prime factors of $2^{t}-1$ must all be of the form $4k+1$, otherwise there would be a prime $p \equiv 1(\bmod 4), p \mid m^{2}+9$, which is a contradiction.

And $2^{t}-1 \equiv-1(\bmod 4)(t>1)$, meaning $-1 \equiv 2^{t}-1 \equiv 1 \cdot 1 \cdot \cdots \cdot 1 \equiv 1(\bmod 4)$, which is a contradiction. Therefore, it must be that $t=1$, i.e., $n=2^{s}$.
The proof below is the same as the previous section's "Skill Training" Exercise 8.