Belarusian Mathematical Olympiad

Answer: $(m,n)=(1,1)$, $(m,n)=(2,5)$. It is easy to see that the pair $(m,n)=(1,1)$ satisfies the condition. For $m>1$ consider the equation $$a^m-b^m=(a-b)(a^{m-1}+a^{m-2}b+a^{m-3}{b}^2+\cdots +a^{m-1}{b}^m).$$

For $a=9$, $b=7$, the first factor is $2$ and the second is greater than $1$. Since all $m$ summands in the second factor are odd numbers, $m$ be even. Let $m=2k$, then $9^{2k}-7^{2k}=(9^k-7^k)(9^k+7^k)=2^n$. Therefore, $9^k-7^k=2^x$, $9^k+7^k=2^y$ where $x<y$ and $x+y=n$. Subtracting the first equation from the second, we get $2\cdot 7^k=2^x(2^{y-x}-1)$, whence $x=1$ and $7^k=2^{y-1}-1$. So $7^k+1$ is divided by $2^{y-1}$. However, if $y\ge 5$ the remainder of $7^k+1$ modulo $16$ (as it is easy to check) is equal to $2$ or $8$, i.e. $7^k+1$ is not divisible by $16$, and therefore there is not such positive integers $k$ for which the pair $(m,n)=(2k,n)$ satisfy the condition $n=y+1\ge 5+1=6$. So $n\le 5$ and $y=n-x=n-1\le 4$, whence $7^k=2^{y-1}-1\le 7$, and thus $k\le 1$. It is easy to see that $m=2k=2\cdot 1=2$ and $n=5$ satisfy the condition. Therefore there are only two pairs $(m,n)=(1,1)$, $(m,n)=(2,5)$ satisfying the problem.