Let $p=x+y+z$, $q=xy+yz+zx$, $r=xyz$. Since $(x+2)(x-2)=x^2-4=y-2$, taking a cyclic product, $\prod_{\rm cyc}(x+2)\prod_{\rm cyc}(x-2)=\prod_{\rm cyc}(x-2)$. If any of $x$, $y$, $z$ equal to $2$, plugging back gives $x=y=z=2$, and so $x+y+z\in\Bbb Z$. Otherwise $1=\prod_{\rm cyc}(x+2)=4p+2q+r+8$, so $4p+2q+r=-7$. Similarly, $\prod_{\rm cyc}(x+1)\prod_{\rm cyc}(x-1)=\prod_{\rm cyc}(x+1)$. If any of $x$, $y$, $z$ equal to $1$, plugging back gives $x=y=z=1$, and so $x+y+z\in\Bbb Z$. Otherwise, $1=\prod_{\rm cyc}(x-1)=r-q+p-1$, so $p-q+r=2$. Directly taking a sum, $x^2+y^2+z^2=x+y+z+6$ gives $p^2-2q=p+6$. By solving the system, $(p,q,r)=(0,-3,-1)$ or $(-1,-2,1)$, and so $p=x+y+z\in\Bbb Z$. We're done.

What a nice question

A trigonometric solution: Using $\cos 2\theta=2\cos^2\theta-1$ we get $x=2\cos\theta\implies y=2\cos 2\theta\implies z=2\cos 4\theta\implies x=2\cos 8\theta$. Therefore $\theta=\frac{2\pi k}{8\pm 1}$ for some $k\in\mathbb{Z}$. This makes $(x,y,z)$ is either $(-1,-1,-1)$ or $(2,2,2)$, or a cycle of either $\left(2\cos\frac{2\pi}{7},2\cos\frac{4\pi}{7},2\cos\frac{6\pi}{7}\right)$ or $\left(2\cos\frac{2\pi}{9},2\cos\frac{4\pi}{9},2\cos\frac{8\pi}{9}\right)$. The tuple $(-1,-1,-1)$ makes $x+y+z=-3$ and $(2,2,2)$ makes $x+y+z=6$. For the other tuples, we have $2\cos\frac{2\pi}{7}+2\cos\frac{4\pi}{7}+2\cos\frac{6\pi}{7}=-1$ because it is the sum of the primitive $7 $th roots of unity and $2\cos\frac{2\pi}{9}+2\cos\frac{4\pi}{9}+2\cos\frac{8\pi}{9}=0$ because it is the sum of the primitive $9 $th roots of unity.