Example 2 Find the positive integer solutions of the equation $3^{x}-5^{y}=z^{2}$. (2009, Balkan Mathematical Olympiad)
Taking modulo 2, we get
$$
0 \equiv z^{2}(\bmod 2) \Rightarrow 21 z \text {. }
$$

Taking modulo 4, we get
$$
(-1)^{x}-1 \equiv 0(\bmod 4) \Rightarrow 2 \mid x \text {. }
$$

Let $x=2 x_{1}$. Then $\left(3^{x_{1}}+z\right)\left(3^{x_{1}}-z\right)=5^{y}$.
Let $\left\{\begin{array}{l}3^{x_{1}}+z=5^{\alpha}, \\ 3^{x_{1}}-z=5^{\beta},\end{array}\right.$ where $\alpha+\beta=y$, and $\alpha>\beta$.
Adding the two equations, we get
$$
2 \times 3^{x_{1}}=5^{\beta}\left(5^{\alpha-\beta}+1\right) \text {. }
$$

Therefore, $\beta=0$. Thus, $\alpha=y$.
(1) If $x_{1}=1$, then $5^{y}+1=6$.
Hence $y=1, x=2 x_{1}=2, z=5^{y}-3^{x_{1}}=2$.
Thus, $(x, y, z)=(2,1,2)$.
(2) If $x_{1} \geqslant 2$, then taking modulo 9, we get
$$
5^{y}+1 \equiv 0(\bmod 9),
$$

i.e., $5^{y} \equiv-1(\bmod 9)$.
However, for any $y \in \mathbf{N}_{+}$, we have
$$
5^{y} \equiv 5,7,-1,4,2,1(\bmod 9) \text {. }
$$

Thus, $y \equiv 3(\bmod 6)$.
Let $y=6 y_{1}+3$. Then
$$
5^{6 y_{1}+3}+1=2 \times 3^{x_{1}} \text {. }
$$

And $5^{6 y_{1}+3}+1=\left(5^{3}\right)^{2 y_{1}+1}+1$ has a factor
$$
5^{3}+1=2 \times 3^{2} \times 7 \text {, }
$$

but $7 \times 2 \times 3^{x_{1}}$, so there is no positive integer solution in this case.
In summary, the positive integer solution to the original equation is
$$
(x, y, z)=(2,1,2) .
$$