Five. (Full marks 14 points) Find the non-negative integer solutions $x, y, z$ that satisfy the equation $2^{x}+3^{y}=z^{2}$.

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If $y=0, 2^{x}=(z-1)(z+1)$, when $z-1=2$, $z=3, z+1=4, x=3$. Thus, we get a solution $(x, y, z)=(3,0,3)$.

If $y>0$, let $x=2a+1$ ($a \geqslant 0$ be an integer), $3^{y}=z^{2}-2^{2a+1}$. Then $z^{2} \equiv 0,1(\bmod 3)$.
$$
2^{2a+1}=\left(2^{2}\right)^{a} \cdot 2^{1} \equiv 1^{a} \cdot 2=2(\bmod 3).
$$

Then $z^{2}-2^{2a+1} \equiv 1$ or $2(\bmod 3)$,
for the corresponding $x$ is not a non-negative integer.
Let $x=2a$ ($a \geq 0$ be an integer), $3^{2}=\left(z-2^{u}\right)\left(z+2^{u}\right)$, then $z-2^{a}=1 \cdot z+2^{a}=3^{x}$. Therefore, $3^{x}=2^{u+1}+1$. Thus, when $a=0$, $(x, y, z)=(0,1,2)$.
When $a \geqslant 1$, $2^{4+1} \equiv 0(\bmod 4)$.
$$
3^{y}=\left\{\begin{array}{l}
3(\bmod 4), \\
1(\bmod 4).
\end{array}\right.
$$

Let $y=2b$ ($b \geqslant 1$ be an integer), $2^{a+1}=3^{2b}-1=\left(3^{b}+1\right) \cdot \left(3^{b}-1\right)$, then $3^{b}-1=2, 3^{b}+1=4$, we get $b=1 \cdot y=2$. Hence, $(x, y, z)=(4,2,5)$.
(Note: By observation alone, 3 sets of solutions are obtained, 1 point for each set of solutions)