1. In the set of natural numbers, solve the equation

$$
3^{x}-5^{y}=z^{2}
$$
1. First, note that $2 \mid z$, so $4 \mid z^{2}=3^{x}-5^{y} \equiv(-1)^{x}-1(\bmod 4)$, from which it follows that $x$ is even, say $x=2 t$. The equation from the problem becomes $\left(3^{t}-z\right)\left(3^{t}+z\right)=5^{y}$, which means that $3^{t}-z=5^{k}$ and $3^{t}+z=5^{y-k}$ for some integer $k \geqslant 0$. But since $5^{k}+5^{y-k}=2 \cdot 3^{t}$ is not divisible by 5, it must be that $k=0$ and

$$
2 \cdot 3^{t}=5^{y}+1
$$

Assume that $t \geqslant 2$. Then $9 \mid 5^{y}+1$, which is true only for $y \equiv 3(\bmod 6)$. However, in that case, $5^{y}+1 \equiv 5^{3}+1 \equiv 0(\bmod 7)$, i.e., $7 \mid 5^{y}+1$, which is impossible.

Therefore, $t \leqslant 1$, and we now obtain the unique solution $(x, y, z)=(2,1,2)$.