Solution:
We have $x^{2} z^{2}=2013\left(x^{2}+x y z+z^{2}\right)$. Let $d=\operatorname{gcd}(x, z)$ and $x=d a, z=d b$. Then $a^{2} b^{2} d^{2}=2013\left(a^{2}+a b y+b^{2}\right)$.
As $\operatorname{gcd}(a, b)=1$, we also have $\operatorname{gcd}\left(a^{2}, a^{2}+a b y+b^{2}\right)=1$ and $\operatorname{gcd}\left(b^{2}, a^{2}+a b y+b^{2}\right)=1$. Therefore $a^{2} \mid 2013$ and $b^{2} \mid 2013$. But $2013=3 \cdot 11 \cdot 61$ is squarefree and therefore $a=1=b$.
Now we have $x=z=d$ and $d^{2}=2013(y+2)$. Once again as 2013 is squarefree, we must have $y+2=2013 n^{2}$ where $n$ is a positive integer.
Hence $(x, y, z)=\left(2013 n, 2013 n^{2}-2,2013 n\right)$ where $n$ is a positive integer.