By checking the cases $n=1,2,3$ we get the solution $n=2$ and $13^{2}=5^{2}+12^{2}$.

If $n=2k+1$ is odd, we consider the equation modulo 5 and we obtain

$$\begin{aligned} x^{2} & \equiv 5^{2k+1} + 12^{2k+1} \pmod{5} \equiv 2^{2k} \cdot 2 \pmod{5} \\ & \equiv (-1)^{k} \cdot 2 \pmod{5} \equiv \pm 2 \pmod{5} \end{aligned}$$

This is not possible, because the square residue of any natural number modulo 5 is 0, 1, or 4. Therefore $n$ is even and $x^{2} = 5^{2k} + 12^{2k}$. Rearrange this equation in the form

$$5^{2k} = (x - 12^k)(x + 12^k)$$

If 5 divides both factors on the right, it must also divide their difference, that is

$$5 \mid (x + 12^k) - (x - 12^k) = 2 \cdot 12^k$$

which is not possible. Therefore we must have

$$x - 12^k = 1 \text{ and } x + 12^k = 5^{2k}$$

By adding the above equalities we get

$$5^{2k} - 1 = 2 \cdot 12^k$$

For $k \geq 2$, we have the inequality

$$25^k - 1 > 24^k = 2^k \cdot 12^k > 2 \cdot 12^k$$

Thus we conclude that there exists a unique solution to our problem, namely $n=2$.