Note that by Jensen's, we have $\sqrt{x+n}+\sqrt{y+n}+\sqrt{z+n}\geq\sqrt{9n+1}$, and by some other methods, we have $\sqrt{x+n}+\sqrt{y+n}+\sqrt{z+n}\leq2\sqrt{n}+\sqrt{n+1}$. Note that $9n+2<(2\sqrt{n}+\sqrt{n+1})^2<9n+3$. Let $\sqrt{x+n}+\sqrt{y+n}+\sqrt{z+n}=m$. We have $9n+1\leq m^2<9n+3$, so either $m^2=9n+1$ or $m^2=9n+2$(but the last possibility is disallowed by taking $\pmod{9}$. So the only integers $n$ that work are those for which $9n+1$ is a perfect square. QED.