IMO Hk TST, 1997

$n$ can be any nonnegative integral power of $2$.

If $n$ has an odd divisor $d \ge 3$, then $2^d - 1 \mid 2^n - 1 \mid m^2 + 81$. Since $2^d - 1 \equiv 3 \pmod{4}$ and $3 \nmid 2^d - 1$, there exists an odd prime $p > 3$ such that $p \equiv 3 \pmod{4}$ and $p \mid 2^d - 1$. This implies $p \mid m^2 + 81$. However, this means $m^2 \equiv -9^2 \pmod{p}$, and hence $(9^{-1}m)^2 \equiv -1 \pmod{p}$. This is impossible as $p \equiv 3 \pmod{4}$. Therefore, $n$ has no odd divisor greater than $1$. Thus, $n = 2^k$ for some nonnegative integer $k$.

It remains to find an $m$ such that $2^{2^k} - 1 \mid m^2 + 81$. Firstly, note that
$$2^{2^k} - 1 = (2 + 1)(2^2 + 1)\cdots(2^{2^{k-1}} + 1).$$
The factors on the right are pairwise relatively prime. Indeed, if $r < s$, then $2^{2^r} + 1 \mid 2^{2^s} - 1$, and $(2^{2^s} - 1, 2^{2^s} + 1) = (2^{2^s} - 1, 2) = 1$, so that $(2^{2^r} + 1, 2^{2^s} + 1) = 1$. Now, by the Chinese remainder theorem, there exists $m \in \mathbb{Z}^+$ such that $3 \mid m$ and
$$m \equiv 9 \cdot 2^{2^{j-1}} \pmod{2^{2^j} + 1}$$
for $j = 1, 2, \dots, k - 1$. For this $m$, we have $3 \mid m^2 + 81$ and
$$m^2 + 81 \equiv 81(2^{2^j} + 1) \equiv 0 \pmod{2^{2^j} + 1}.$$
Hence, $2^{2^k} - 1 \mid m^2 + 81$ as desired.