Example 6 Try to find the non-negative integer solutions $x, y, z$ that satisfy the equation $2^{x}+3^{y}=z^{2}$.
If $y=0$, then
$2^{x}=z^{2}-1=(z+1)(z-1)$.
Let $\left\{\begin{array}{l}z+1=2^{s}, \\ z-1=2^{t}\end{array}\right.$. $(s>t, x=s+t)$.
Subtracting the two equations gives $2^{s}-2^{t}=2$.
Solving this, we get $t=1, s=2$.
At this point, $x=3, y=0, z=3$.
If $y>0$, then
$$
2^{x}+3^{y} \equiv(-1)^{x}(\bmod 3) \text {. }
$$

If $x$ is odd, then
$$
z^{2} \equiv 2^{x}+3^{y} \equiv 2(\bmod 3),
$$

which is impossible.
Therefore, $x$ is even, let $x=2a$.
Then $3^{y}=z^{2}-2^{2a}=\left(z+2^{a}\right)\left(z-2^{a}\right)$.
Let $\left\{\begin{array}{l}z+2^{a}=3^{s}(s \geqslant 1), \\ z-2^{a}=3^{t}(t \geqslant 0) .\end{array}\right.$
Subtracting the two equations gives $2^{a+1}=3^{s}-3^{t}$.
Solving this, we get $t=0$. Hence $2^{a+1}=3^{s}-1$.
If $a=0$, then $s=1$.
At this point, $x=0, y=1, z=2$.
If $a>0$, then
$$
2^{a+1} \equiv 3^{s}-1 \equiv(-1)^{s}-1 \equiv 0(\bmod 4) \text {. }
$$

At this point, $s$ must be even, let $s=2b$.
Then $2^{a+1}=3^{2b}-1=\left(3^{b}+1\right)\left(3^{b}-1\right)$.
Let $\left\{\begin{array}{l}3^{b}+1=2^{m}, \\ 3^{b}-1=2^{n}\end{array},(m+n=a+1)\right.$.
Subtracting the two equations gives $2^{m}-2^{n}=2$.
Solving this, we get $m=2, n=1$.
At this point, $x=4, y=2, z=5$.
In summary, the original equation has three solutions.