Note that since expressions are symmetric, if $(x,y,z)$ is a solution, then any permutation of it is also a solution. First condition gives $(x-y)(y-z)(z-x) = 0$. So WLOG assume $y = z$. Second becomes $(x-y)^2 = 4 \implies x-y \in \{-2,2\}$. So all solutions are $(x,x-2,x-2)$ (upto permutations and $x \in \mathbb{R} - \{0,2\}$) and $(x,x+2,x+2)$ (upto permutations and $x \in \mathbb{R} -\{0,-2\}$)