Example 4 Find all positive integer triples $(x, y, z)$ such that $y$ is a prime number, and neither 3 nor $y$ is a divisor of $z$, and satisfying $x^{3}-y^{3}=z^{2}$.

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From the original equation, we get $z^{2}=(x-y)\left(x^{2}+x y+z^{2}\right)$.
Let $\left(x-y, x^{2}+x y+y^{2}\right)=\left(x-y,(x-y)^{2}+3 x y\right)=(x-y, 3 x y)=\left(x-y, 3 y^{2}\right)=d$, then $d \mid 3 y^{2}$. Since $y$ is a prime and 3 and $y$ are not divisors of $z$, we have $d=1$. Therefore, $x-y$ and $x^{2}+x y+y^{2}$ are both perfect squares. We set
$$
\left\{\begin{array}{l}
x-y=u^{2}, \\
x^{2}+x y+y^{2}=v^{2},
\end{array}\right.
$$

where $u, v \in \mathbf{N}^{*}$. Multiplying both sides of equation (2) by 4, completing the square, rearranging, and factoring, we get
$$
(2 v-2 x-y)(2 v+2 x+y)=3 y^{2} \text {, }
$$

Noting that $y$ is a prime, and $2 v+2 x+y \in \mathbf{N}^{*}$, and $2 v-2 x-y<2 v+2 x+y$, we have
$$
\left\{\begin{array}{l}
2 v-2 x-y=1, y, 3, \\
2 v+2 x+y=3 y^{2}, 3 y, y^{2} .
\end{array}\right.
$$

For the first case, subtracting the two equations, we get $3 y^{2}-1=2(2 x+y)=4 u^{2}+6 y$.
Thus, $u^{2}+1 \equiv 0(\bmod 3)$, but $u^{2} \equiv 0$ or $1(\bmod 3)$, so there is no solution in this case.
For the second case, we have $2 y=4 x+2 y$, leading to $x=0$, which also has no solution.
For the third case, we have $y^{2}-3=4 u^{2}+6 y \Rightarrow(y-2 u-3)(y+2 u-3)=12$,
solving which gives $(y, u)=(7,1)$, and thus $(x, z)=(8,13)$.
Therefore, the positive integer solution to the original equation is $(x, y, z)=(8,7,13)$.