HMMT February

Solution:
First consider when $n \geq m$, so let $n = m + d$ where $d \geq 0$. Then we have
$$2^{m}(m + d - 2^{d} m) = 2^{m}(m(1 - 2^{d}) + d)$$
which is non-positive unless $m = 0$. So our first set of solutions is $m = 0, n = 2^{j}$.

Now, we can assume that $m > n$, so let $m = n + d$ where $d > 0$. Rewrite
$$2^{m} n - 2^{n} m = 2^{n + d} n - 2^{n}(n + d) = 2^{n}((2^{d} - 1)n - d)$$
In order for this to be a power of $2$, $(2^{d} - 1)n - d$ must be a power of $2$. This implies that for some $j$, $2^{j} \equiv -d \pmod{2^{d} - 1}$. But notice that the powers of $2$ modulo $2^{d} - 1$ are $1, 2, 4, \ldots, 2^{d-1}$ (since $2^{d} \equiv 1$ so the cycle repeats).

In order for the residues to match, we need $2^{j} + d = c(2^{d} - 1)$, where $0 \leq j \leq d-1$ and $c \geq 1$. In order for this to be true, we must have $2^{d-1} + d \geq 2^{d} - 1 \Longleftrightarrow d + 1 \geq 2^{d-1}$. This inequality is only true for $d = 1, 2, 3$. We plug each of these into the original expression $(2^{d} - 1)n - d$.

For $d = 1$: $n - 1$ is a power of $2$. This yields the set of solutions $(2^{j} + 2, 2^{j} + 1)$ for $j \geq 0$.

For $d = 2$: $3n - 2$ is a power of $2$. Note that powers of $2$ are $-2 \pmod{3}$ if and only if it is an even power, so $n = \frac{2^{2j} + 2}{3}$. This yields the solution set $\left(\frac{2^{2j} + 8}{3}, \frac{2^{2j} + 2}{3}\right)$, $j \geq 0$.

For $d = 3$: $7n - 3$ is a power of $2$. Powers of $2$ have a period of $3$ when taken $\pmod{7}$, so inspection tells us $7n - 3 = 2^{3j + 2}$, yielding the solution set $\left(\frac{2^{3j + 2} + 24}{7}, \frac{2^{3j + 2} + 3}{7}\right)$, $j \geq 0$.

Therefore, all the solutions are of the form
$$\begin{gathered} (m, n) = (0, 2^{j}),\ (2^{j} + 2, 2^{j} + 1) \\ \left(\frac{2^{2j} + 8}{3}, \frac{2^{2j} + 2}{3}\right),\ \left(\frac{2^{3j + 2} + 24}{7}, \frac{2^{3j + 2} + 3}{7}\right) \end{gathered}$$
for $j \geq 0$.

Restricting this family to $m, n \leq 100$ gives $7 + 7 + 5 + 3 = 22$.