Saudi Arabian Mathematical Competitions, 2015

Notice that $14^{x} \geq 2015$ which implies that $x \geq 3$. On the other hand $3^{y} = 14^{x} - 2015 \geq 14^{3} - 2015 = 729 = 3^{6}$. We deduce that $y \geq 6$.

We have $3^{y} = 14^{x} - 2015 \equiv 1 \pmod{7}$. But $\operatorname{Ord}_{7} 3 = 6$. We deduce that $y = 6k$ for some positive integer $k$.

We have $0 \equiv 2015 \equiv 14^{x} - 3^{6k} \equiv (-1)^{x} - (-1)^{k} \pmod{5}$. We deduce that $x$ and $k$ have the same parity.

Because $y \geq 6$, we have $5^{x} \equiv 14^{x} - 3^{y} \equiv 2015 \equiv -1 \pmod{9}$. This implies that $x \equiv 3 \pmod{6}$ and therefore both $x$ and $k$ are odd numbers.

Now, assume that $x > 3$. This means that $14^{x} \equiv 0 \pmod{16}$ which implies that $9^{k} \equiv 3^{6k} \equiv 1 \pmod{16}$. But $\operatorname{Ord}_{16} 9 = 2$, which means that $k$ is even, which is a contradiction.

We deduce that $x = 3$ and hence $y = 6$ which gives the unique solution of the equation.