22nd Korean Mathematical Olympiad Final Round

Since $3^m-2\equiv 0(\mathrm{mod}7)$, one may easily show that $m\equiv 2(\mathrm{mod}6)$. If $m=2$ then $n=1$, which is a solution of the equation.

Assume that $m=2s\ge 4$. Note that $2+7^n$ is divisible by $27$ in this case and $$7^9\equiv 1(\mathrm{mod}27),7^1\equiv -2(\mathrm{mod}27).$$ Hence we have $n\equiv 4(\mathrm{mod}9)$. If we let $n=9t+4$, then $$2+7^{9t+4}\equiv 2+7^4\equiv 35(\mathrm{mod}37).$$ For any positive integer $u$ less than $10$, $$9^n\equiv 9,7,26,12,34,10,16,33,1(\mathrm{mod}37).$$ Therefore there does not exist a solution for any $m\ge 4$. $\square$