SAMC

We can write $$x^2+y^2+z^2+1=xy+yz+zx+|x-y+z-y|$$ hence $$(x-y{)}^2+(y-z{)}^2+(z-x{)}^2+2=2|x-y+z-y|.$$ It follows $$(x-y{)}^2+(y-z{)}^2+(z-x{)}^2+2\le 2|x-y|+2|y-z|.$$ The last relation is equivalent to $$(|x-y|-1{)}^2+(|y-z|-1{)}^2+(z-x{)}^2\le 0.$$ We get $|x-y|=1$, $|y-z|=1$ and $x=z$. The desired triples $(x,y,z)$ are $(a,a-1,a)$, $(a,a+1,a)$, where $a\in \mathbb{R}$.