First consider when $n \geq m$, so let $n=m+d$ where $d \geq 0$. Then we have $2^{m}\left(m+d-2^{d} m\right)=$ $2^{m}\left(m\left(1-2^{d}\right)+d\right)$, 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 $2^{m} n-2^{n} m=2^{n+d} n-$ $2^{n}(n+d)=2^{n}\left(\left(2^{d}-1\right) n-d\right)$. In order for this to be a power of $2,\left(2^{d}-1\right) n-d$ must be a power of 2. This implies that for some $j, 2^{j} \equiv-d\left(\bmod 2^{d}-1\right)$. But notice that the powers of $2\left(\bmod 2^{d}-1\right)$ are $1,2,4, \ldots, 2^{d-1}\left(2^{d} \equiv 1\right.$ so the cycle repeats $)$. In order for the residues to match, we need $2^{j}+d=c\left(2^{d}-1\right)$, 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 $\left(2^{d}-1\right) n-d$. For $d=1: n-1$ is a power of 2 . This yields the set of solutions $\left(2^{j}+2,2^{j}+1\right)$ for $j \geq 0$. For $d=2: 3 n-2$ is a power of 2 . Note that powers of 2 are $-2(\bmod 3)$ if and only if it is an even power, so $n=\frac{2^{2 j}+2}{3}$. This yields the solution set $\left(\frac{2^{2 j}+8}{3}, \frac{2^{2 j}+2}{3}\right), j \geq 0$. For $d=3: 7 n-3$ is a power of 2 . Powers of 2 have a period of 3 when taken $(\bmod 7)$, so inspection tells us $7 n-3=2^{3 j+2}$, yielding the solution set $\left(\frac{2^{3 j+2}+24}{7}, \frac{2^{3 j+2}+3}{7}\right), j \geq 0$. Therefore, all the solutions are of the form $$\begin{gathered} (m, n)=\left(0,2^{j}\right),\left(2^{j}+2,2^{j}+1\right) \\ \left(\frac{2^{2 j}+8}{3}, \frac{2^{2 j}+2}{3}\right),\left(\frac{2^{3 j+2}+24}{7}, \frac{2^{3 j+2}+3}{7}\right) \end{gathered}$$ for $j \geq 0$. Restricting this family to $m, n \leq 100$ gives $7+7+5+3=22$.