SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Considering the equation $\mathrm{mod}5$ we get $$3^m\equiv -1(\mathrm{mod}5),$$ so $m=4k+2$ for some positive integer $k$. Then, considering the equation $\mathrm{mod}7$ we get $$-2^n-9^{2k+1}\equiv 5(\mathrm{mod}7),$$ which means $$2^n+2^{2k+1}\equiv 2(\mathrm{mod}7).$$ Since $2^s\equiv 1,2,4(\mathrm{mod}7)$ so the only possibility is $2^n\equiv 2^{2k+1}\equiv 1(\mathrm{mod}7)$ so $3\mid n$ and $3\mid 2k+1$. From the last one we get $3\mid m$ so we can write $n=3x$ and $m=3y$. Therefore, the given equation takes the form $5^3\cdot 2^{3x}-3^{3y}=271$, or $$\left(5\cdot 2^x-3^y\right)\left(25\cdot 2^{2x}+5\cdot 2^x\cdot 3^y+3^{2y}\right)=271.$$ It follows that $25\cdot 2^{2x}+5\cdot 2^x\cdot 3^y+3^{2y}\le 271$ and so $x<2$. We conclude $x=1$ and then $y=2$. So $m=6$ and $n=3$. $\square$