Solution:
We seek solutions $(x, y, z)$ which are in arithmetic progression. Let us put $y-x=z-y=d>0$ so that the equation reduces to the form
$$3 y^{2}+2 d^{2}=2 d^{3}$$
Thus we get $3 y^{2}=2(d-1) d^{2}$. We conclude that $2(d-1)$ is 3 times a square. This is satisfied if $d-1=6 n^{2}$ for some $n$. Thus $d=6 n^{2}+1$ and $3 y^{2}=d^{2} \cdot 2\left(6 n^{2}\right)$ giving us $y^{2}=4 d^{2} n^{2}$. Thus we can take $y=2 d n=2 n\left(6 n^{2}+1\right)$. From this we obtain $x=y-d=(2 n-1)\left(6 n^{2}+1\right), z=y+d=(2 n+1)\left(6 n^{2}+1\right)$. It is easily verified that
$$(x, y, z)=\left((2 n-1)\left(6 n^{2}+1\right), 2 n\left(6 n^{2}+1\right),(2 n+1)\left(6 n^{2}+1\right)\right)$$
is indeed a solution for a fixed $n$ and this gives an infinite set of solutions as $n$ varies over natural numbers.