This is more of a hint and general approach:
WLOG we can assume $x,y,z \geq 0$ - otherwise consider $(-x)$ etc.
Note that none of them can equal $0$.
We have $x \geq 2, y \geq 2 , z \geq 3$ by using that $x,y,z \geq 1$
If $x,y,z \geq 4$ then plugging in we get a contradiction. Hence at least one of the $3$ is $\leq 3$
We can now consider again a case by case approach:
Case $1$: $x=2$
Case $2$: $x=3$
Case $3$: $y=2$
Case $4$: $y=3$
Case $5$: $z=3$
Then repeat the idea of finding an upper bound to get all solutions. Once you've done that go back to the WLOG bit and say that if $(x,y,z)$ is a solution then we can make $7$ more solutions by just changing signs