Example 6 In the set of integers, find all solutions to the equation $2^{x}+3^{y}=z^{2}$.

 

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Solution: Since $2^{x}$ is even and $3^{y}$ is odd, therefore, $z^{2}$ is odd, which means $z$ is odd, and $3 x_{z}$, so we can let $z=3 k \pm 1$, thus $z^{2}=(3 k \pm 1)^{2} \equiv 1(\bmod 3)$.

Consider the parity of $x$. If $x=2 n, n \in Z$, then $2^{x} \equiv 1(\bmod 3)$; if $x=2 n+1, n \in Z$, then $2^{x}=2^{2 n+1} \equiv 2(\bmod 3)$. Therefore, only when $x$ is even, the remainders of both sides of the equation modulo 3 are the same.

Let $x=2 n$, the equation can be transformed into $4^{n}+3^{y}=z^{2}$, and since $z$ is odd, $z^{2} \equiv 1(\bmod 4)$.

Now consider the parity of $y$. If $y=2 m+1, m \in Z$, then $3^{y} \equiv 3(\bmod 4)$; if $y=2 m, m \in Z$, then $3^{y} \equiv 1(\bmod 4)$, therefore, $y$ must be even.

Let $y=2 m$, then the original equation can be transformed into $4^{n}+9^{m}=z^{2}$, i.e., $2^{2 n}=\left(z-3^{m}\right)(z+3^{m})$. Set $z-3^{m}=2^{a}, z+3^{m}=2^{b}$ (where $a+b=2 n, a<b$), then $z=2^{a-1}(1+2^{b-a})$.
Since $z$ is odd, so $a=1, b=2 n-1, z=1+2^{x-2}$.
Thus $3^{m}=2^{x-2}-1=4^{n-1}-1 \equiv 3(\bmod 4)$, from which we know $m$ is odd, hence $4^{n-1}=3^{m}+1=4\left(3^{m-1}-3^{m-2}+\cdots+1\right)$.

Since $3^{m-1}-3^{m-2}+\cdots+1$ is the sum of an odd number of odd numbers, its sum is odd, so $3^{m-1}-3^{m-2}+\cdots+1=1$. This gives $n=2, x=4, m=1$, from which we deduce $y=2 m=2, z= \pm 5$.
Therefore, the solutions of the original equation in the set of integers are $(4,2,5)$ and $(4,2,-5)$.