Four, (50 points) Find all integer solutions $(x, y, z)$ of the equation $2^{x}+3^{y}=z^{2}$.

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(4) Solution: From the equation $2^{x}+3^{y}=z^{2}$ (1), the left side is an integer and the right side is a positive integer, so: $x \geqslant 0, y \geqslant 0, z \neq 0$.
If $(x, y, z)$ is an integer solution to (1), then $(x, y, -z)$ is also an integer solution to (1); conversely, the same holds. Therefore, to find all integer solutions to (1), it is only necessary to find those solutions to (1) where $z>0$. Hereafter, we always assume that $x, y$ are non-negative integers, and $z$ is a positive integer.
(1) When $x=0$, from (1) we have $1+3^{y}=z^{2}$, thus $3^{y}=(z-1)(z+1)$. Since 3 is a prime number, $z-1$ and $z+1$ are powers of 3. Let $z-1=3^{t}, z+1=3^{t}$. Here $s, t$ are non-negative integers, $s<t$, and $s+t=y$.
Since $3^{t}-3^{s}=2$, we have $3^{s}(3^{t-s}-1)=2$.
Since 2 is a prime number and $3 \neq 2$, we have $3^{s}=1$ and $3^{t-s}-1=2$, which gives $s=0, t=1$, and thus $y=1, z=2$.
Therefore, when $x=0$, (1) has and only has the integer solution $(0,1,2)$.
(2) When $y=0$, (1) becomes $2^{x}+1=z^{2}$, thus $2^{x}=(z+1)(z-1)$.
Similarly to (1), we get $x=3, z=3$. At this point, (1) has and only has the integer solution $(3,0,3)$.
(3) When $x, y$ are both positive integers, $x, y$ can only be even numbers, because:
Assume $x$ is odd, let $x=2p+1, p$ be a non-negative integer. From (1) we have $2^{2p+1}+3^{y}=z^{2}$, i.e., $2(3+1)^{p}+3^{y}=z^{2}$, and $2(3+1)^{p}+3^{y} \equiv 2 \pmod{3}$, while $z^{2} \equiv 0$ or $1 \pmod{3}$, which is a contradiction. Therefore, $x$ is even, let $x=2p, p$ be a positive integer.

Assume $y$ is odd, let $y=2q+1, q$ be a non-negative integer, from (1) we have $4^{p}+3(8+1)^{q}=z^{2}$, and $4^{p}+3(8+1)^{q} \equiv 3 \pmod{4}$, while $z^{2} \equiv 0$ or $1 \pmod{4}$, which is a contradiction. Therefore, $y$ is even, so $y=2q, q \in \mathbf{N}^{*}$.

Thus (1) becomes $2^{2p}+3^{2q}=z^{2}$, i.e., $3^{2q}=(z-2^{p})(z+2^{p})$. From this equation, we know that $z-2^{p}$ and $z+2^{p}$ are both powers of 3, their difference is $(z+2^{p})-(z-2^{p})=2^{p+1}$, so $z-2^{p}=1$, otherwise $3 \mid 2^{p+1}$, which is impossible (since 2 and 3 are coprime). Thus we get $z=2^{p}+1$ and substitute it into (1) to get $2^{2p}=(1+2^{p}-3^{q})(1+2^{p}+3^{q})$.
Thus we can set $1+2^{p}-3^{q}=2^{u}, 1+2^{p}+3^{q}=2^{v}$, where $u, v$ are non-negative integers, $0 \leqslant u<v$ and $u+v=2p$.
Since $2^{v}-2^{u}=2^{u}(2^{v-u}-1)=2 \cdot 3^{q}$, so $2^{u}=1$ or 2.
Assume $2^{u}=1$, i.e., $1+2^{p}-3^{q}=1$, then $2^{p}=3^{q}$, which is a contradiction. Therefore, $2^{u}=2$. Thus $2^{p}=3^{q}+1$.
Assume $p$ is odd, then the left side of the above equation is 2 modulo 3, while the right side is 1 modulo 3, which is a contradiction. Therefore, $p$ is even.
Let $p=2b, b \in \mathbf{N}^{*}$, the above equation can be written as $1+3^{q}=(2^{b})^{2}$. From (1) we know $q=1, 2^{b}=2, b=1$,
Thus $x=2p=4b=4, y=2q=2, z=5$. Therefore, when $x, y \neq 0$, (1) has and only has the integer solution $(4,2,5)$.
In summary, all integer solutions to (1) are: $(0,1,2),(0,1,-2),(3,0,3),(3,0,-3),(4,2,5),(4,2,-5)$.