Taiwan IMO Selection Camp

(6, 3), (9, 3), (9, 5) and (54, 5).

Note that when $n$ is fixed, this is a quadratic equation in $m$; the condition for $m$ to have an integer solution is that the discriminant is a perfect square. Solving the quadratic equations for $n = 0, 1, 2, 3, 4, 5$ shows that for $n \le 5$ there are exactly the four solutions described above. Below we prove that there are no solutions for $n \ge 6$.

Suppose $(m, n)$ satisfies the equation of this problem and $n \ge 6$, then $m|2 \cdot 3^n = m(2^{n+1}-m-1)$,
so $m$ can be written as $m = 3^p$ or $m = 2 \cdot 3^q$.

In the first case, let $q = n - p$, we have
$$2^{n+1} - 1 = m + \frac{2 \cdot 3^n}{m} = 3^p + 2 \cdot 3^q.$$
In the second case, let $p = n - q$, we have
$$2^{n+1} - 1 = m + \frac{2 \cdot 3^n}{m} = 2 \cdot 3^q + 3^p.$$
Thus in either case we have $2^{n+1} - 1 = 3^p + 2 \cdot 3^q$, with $0 \le p, q \le n, p + q = n$.

Now we estimate the range of $p, q$, consider
$$3^p < 2^{n+1} = 8^{\frac{n+1}{3}} < 9^{\frac{n+1}{3}} = 3^{\frac{2(n+1)}{3}}$$
Similarly we have $2 \cdot 3^q < 3^{\frac{2(n+1)}{3}}$, so $p, q < \frac{2(n+1)}{3}$. From $p+q = n$ we also obtain $p, q > \frac{n-2}{3}$.

Let $h = \min(p, q)$, we have $3^h|3^p+2 \cdot 3^q = 2^{n+1}-1$. Then since $h > \frac{n-2}{3} > 1$,
$2^{n+1}-1$ is a multiple of 9. It is easy to deduce that in this case $6|n+1$.

Therefore we may write $n + 1 = 6r$. Then we have
$$2^{n+1} - 1 = 4^{3r} - 1 = (4^{2r} + 4^r + 1)(2^r + 1)(2^r - 1).$$
Note that $4^{2r} + 4^r + 1 = (4^r - 1)^2 + 3 \cdot 4^r$ is certainly a multiple of 3 but certainly not a multiple of 9. Also $2^r + 1$ and $2^r - 1$ differ by 2, so only one of them can be a multiple of 3. Whichever one it is, it must be divisible by $3^{h-1}$. In either case, we have
$$3^{h-1} \le 2^r + 1 \le 3^r$$
hence $h - 1 \le r$. But earlier we had $h > \frac{n-2}{3}$, and $r = \frac{n+1}{6}$, from which it clearly follows that $n < 11$. This contradicts $6|n+1$ and $n \ge 6$, so there are no solutions for $n \ge 6$.