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.