XIV OBM

We must have $n>1$, $c>a$ and $c>b$. Suppose without loss of generality that $a\le b$. Dividing by $n^a$ we get $1+n^{b-a}=n^{c-a}$. But $c-a>b-a$, so $n^{c-a}\ge n\cdot n^{b-a}$. So we must have $n=2$, $b-a=0$ and $c-a=1$. It is easy to check that that gives a solution.