Example 11 Find all non-negative integers $x, y, z$, such that
$$2^{x}+3^{y}=z^{2}$$
Solve (1) When $y=0$, we have
$$2^{x}=z^{2}-1=(z-1)(z+1)$$

Thus, we can set
$$z-1=2^{\alpha}, z+1=2^{\beta}, 0 \leqslant \alpha \leqslant \beta$$

Therefore,
$$2^{\beta}-2^{\alpha}=2$$

At this point, if $\alpha \geqslant 2$, then $4 \mid 2^{\beta}-2^{\alpha}$, which contradicts $4 \nmid 2$, so $\alpha \leqslant 1$. And $\alpha=0$ leads to $2^{\beta}=$
3, which is a contradiction, hence

So
$$\begin{array}{l}
\alpha=1, \beta=2 \\
z=3, x=3
\end{array}$$

We get
$$(x, y, z)=(3,0,3)$$
(2) When $y>0$, since $3 \nmid 2^{x}+3^{y}$, it follows that $3 \nmid z$, so
$$z^{2} \equiv 1(\bmod 3)$$

Taking (1) modulo 3, we know
$$(-1)^{x} \equiv 1(\bmod 3)$$

Thus, $x$ is even. Now set $x=2 m$, then
$$\left(z-2^{m}\right)\left(z+2^{m}\right)=3^{y}$$

So we can set
$$z-2^{m}=3^{\alpha}, z+2^{m}=3^{\beta}, 0 \leqslant \alpha \leqslant \beta, \alpha+\beta=y$$

Thus,
$$3^{\beta}-3^{\alpha}=2^{m+1},$$

If $\alpha \geqslant 1$, then $3 \mid 3^{\beta}-3^{\alpha}$, but $3 \nmid 2^{m+1}$, which is a contradiction, so $\alpha=0$, hence
$$3^{\beta}-1=2^{m+1}$$

When $m=0$, $\beta=1$, we get
$$(x, y, z)=(0,1,2)$$

When $m>0$, $2^{m+1} \equiv 0(\bmod 4)$, so
$$3^{\beta} \equiv 1(\bmod 4)$$

This requires $\beta$ to be even. Set $\beta=2 n$, then
$$2^{m+1}=3^{2 n}-1=\left(3^{n}-1\right)\left(3^{n}+1\right)$$

As in the case when $y=0$, we know
$$3^{n}-1=2$$

Thus $n=1$, and $m=2$, we get
$$(x, y, z)=(4,2,5)$$

Therefore, $(x, y, z)=(3,0,3),(0,1,2),(4,2,5)$.