Solution:
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 $n=2$.