18 4 \% 6 Solve the equation for positive integer solutions:
$$
5^{x}-3^{y}=2
$$
The solution $x=y=1$ is clearly a solution to (1).
Assume $x, y$ are not both 1, then neither of them is 1. Taking (1) modulo 4, we get $1-(-1)^{y}=2(\bmod 4)$, so $y$ is odd and greater than 1. Taking (1) modulo 9, we get $5^{x} \equiv 2$ (mod 9), so $x=6k+5$. Finally, taking (1) modulo 7, we have $3^{y} \equiv 3,-1,-2(\bmod 7)$, and $5^{x} \equiv 5^{5} \equiv 2(\bmod 7)$, thus $5^{x}-3^{y} \equiv-1,3,4(\bmod 7)$, meaning (1) does not hold. Therefore, the only solution to this problem is $x=y=1$.