From the problem, we have
$$(x-y)\left[(x-y)^{2}+3 x y\right]=z^{2} \text {. }$$

Since $y$ is a prime number, and $y \nmid z, 3 \nmid z$, combining with equation (1), we know
$$\begin{array}{c} (x, y)=1, (x-y, 3)=1 \text {. Then } \\ \left(x^{2}+x y+y^{2}, x-y\right) \\ =(3 x y, x-y)=1 . \end{array}$$

From equations (1) and (2), we get
$$x-y=m^{2}, x^{2}+x y+y^{2}=n^{2}, z=m n \text {. }$$
$\left(m, n \in \mathbf{N}_{+}\right)$
Thus, $3 y^{2}=4 n^{2}-(2 x+y)^{2}$
$$=(2 n+2 x+y)(2 n-2 x-y) \text {. }$$

Since $y$ is a prime number, and $2 n-2 x-y < 2 n+2 x+y$, there are the following three cases.
(1) $2 n-2 x-y=y, 2 n+2 x+y=3 y$, which gives $x=0$, discard.
(2) $2 n-2 x-y=3, 2 n+2 x+y=y^{2}$, then
$$\begin{array}{l} y^{2}-3=4 x+2 y \\ =4\left(m^{2}+y\right)+2 y=4 m^{2}+6 y, \end{array}$$

which simplifies to $(y-3)^{2}-4 m^{2}=12$.
Solving this, we get $y=7, m=1$.
Therefore, $x=8, y=7, z=13$.
(3) $2 n-2 x-y=1, 2 n+2 x+y=3 y^{2}$, then
$$\begin{array}{l} 3 y^{2}-1=4 x+2 y \\ =4\left(m^{2}+y\right)+2 y=2\left(2 m^{2}+3 y\right), \end{array}$$

which simplifies to $3 y^{2}-6 y-3 m^{2}=m^{2}+1$.
Thus, $m^{2}+1=0(\bmod 3)$, which contradicts $m^{2} \equiv 0,1(\bmod 3)$.

In summary, the positive integer solution that satisfies the conditions is unique, which is $(8,7,13)$.