A very routine problem, to get you started: Assume $x\ge y\ge z$ so that $3xy \ge xy+yz+zx=xyz+2$. So $3>z$.

Continuing, we next proceed by casework. Case 1: $z=2$ $\implies 2x+2y-xy=2 \implies (x-2)(2-y)=-2 \implies x=4, y=3$ Case 2: $z=1$ $\implies x+y=2 \implies x=y=1$. Thus, the solutions are $(x,y,z)\in\{(1,1,1), (4,3,2)\}$ and permutations.

If i'm not wrong, z can be equal to 0 too, so we also obtain the solution (0,1,2) and its permutations

@tudordarius it is given that $x,y,z\in \mathbb{Z_+}$, so none of them can be equal to $0$.

tudordarius wrote: If i'm not wrong, z can be equal to 0 too, so we also obtain the solution (0,1,2) and its permutations You're wrong, it says positive integers

You can also rearrange the equation as $(x-1)(y-1)(z-1)=x+y+z-3$, which is equivalent to solving $abc=a+b+c$.

chess12500 wrote: You can also rearrange the equation as $(x-1)(y-1)(z-1)=x+y+z-3$, which is equivalent to solving $abc=a+b+c$. Continuing, WLOG $a\geq b\geq c$, then $abc=a+b+c\leq 3a$ so $bc\leq 3$, which only occurs if $(b,c)={(1,3), (0,3), (1,2), (0,2), (1,1), (0,1), (0,0)}$ and permutations. Casework now finishes it off. Specifically, for b=1, c=3, we have x,y,z are in some order 2,4,3 (3 comes from a=2 since 3a=4+a), etc.

huashiliao2020 wrote: WLOG $a\geq b\geq c$ then $c^3\leq abc=a+b+c\leq 3c$, so $c\leq 1$ This part is wrong. we have $a+b+c\ge 3c$ not $a+b+c\le 3c$.

Sorry. Fixed.

WLOG $x \ge y \ge z$ so $3xy \ge xyz+2$ so $z \le 2$ Case 1 (if $z=2$): $2x+2y-xy=2$, by SFFT $(x-2)(y-2)=2$, so $x=4$ and $y=3.$ Case 2 (if $z=1$): $x+y=2$, so $x=y=1.$ Hence, the only solutions are $\boxed{(1,1,1), (4,3,2)}$, and permutations.