Solution:
If $z=0$, then $x=0$ and $y=0$, and $(x, y, z)=(0,0,0)$.

Let us assume that $z \neq 0$, and $x+y=a$ and $x-y=b$ where $a$ and $b$ are nonzero integers such that $z=a b$. Then $x=(a+b)/2$ and $y=(a-b)/2$, and the second equation gives $3 a^{2}-3 b^{2}+4 a b^{2}=4 a^{2} b^{2}$.

Hence
$$b^{2}=\frac{3 a^{2}}{4 a^{2}-4 a+3}$$
and
$$3 a^{2} \geq 4 a^{2}-4 a+3$$
which is satisfied only if $a=1,2$ or $3$.

- If $a=1$, then $b^{2}=1$. $(x, y, z)=(1,0,1)$ and $(0,1,-1)$ are the only solutions in this case.
- If $a=2$, then $b^{2}=12/11$. There are no solutions in this case.
- If $a=3$, then $b^{2}=1$. $(x, y, z)=(1,2,-3)$ and $(2,1,3)$ are the only solutions in this case.

To summarize, $(x, y, z)=(0,0,0), (1,0,1), (0,1,-1), (1,2,-3)$ and $(2,1,3)$ are the only solutions.