5. Proof: The equation clearly has a solution $x=y=1$. Taking the equation modulo 4, it is easy to see that $y$ is odd. If $y>1$, taking the equation modulo 9 yields
$$5^{x} \equiv 2(\bmod 9)$$

It is not difficult to find that for $x=1,2, \cdots, 5^{x}$ modulo 9, the cycle is $5,7,8,4,2,1$. Therefore, by (1), $x$ must be of the form $6k+5$. Taking the original equation modulo 7, it is easy to verify that for odd $y$, $3^{y} \equiv 3,5,6(\bmod 7)$.

When $x=6k+5$, by Fermat's Little Theorem, $5^{6} \equiv 1(\bmod 7)$, so
$$5^{x}=5^{6k+5} \equiv 5^{5} \equiv 3(\bmod 7)$$

Thus, the two sides of the original equation are not congruent modulo 7, so it has no solutions for $y>1$. Therefore, the only positive integer solution is $y=1, x=1$.