Example 1 Find all positive integer triples $(x, y, z)$ such that $y$ is a prime, $y$ and 3 are not divisible by $z$, and $x^{3}-y^{3}=$ $z^{2}$.
Analyze the factorization of $x^{3}-y^{3}=z^{2}$, using the fact that $y$ is a prime number, we get $x-y=m^{2}, x^{2}+x y+y^{2}=n^{2}, z=m n$, thus we have $3 y^{2}=(2 n+2 x+y)(2 n-2 x-y)$. Then, by using the fact that $y$ is a prime number, we can discuss the cases.

Solution: From the problem, we know $(x-y)\left[(x-y)^{2}+3 x y\right]=z^{2}$,
Since $y$ is a prime number, and $y$ and 3 are not divisors of $z$, we have $(x, y)=1, (x-y, 3)=1$.
Thus, $\left(x^{2}+y^{2}+x y, x-y\right)=(3 x y, x-y)=1$,
From (1) and (2), we get
$x-y=m^{2}, x^{2}+x y+y^{2}=n^{2}, z=m n\left(m, n \in \mathbf{N}_{+}\right)$, ?
Therefore, $3 y^{2}=4 n^{2}-(2 x+y)^{2}=(2 n+2 x+y)(2 n-2 x-y)$.
Since $y$ is a prime number, and $2 n-2 x-y < 2 n+2 x+y$,
There are the following 3 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 $y^{2}-3=4 x+2 y=4 m^{2}+6 y$,
So $(y-3)^{2}-4 m^{2}=12$, solving gives $y=7, m=1$.
Thus, $x=8, y=7, z=13$.
(3) $2 n-2 x-y=1$, $2 n+2 x+y=3 y^{2}$, then $3 y^{2}-1=4 x+2 y=2\left(2 m^{2}+3 y\right)$,
That is, $3 y^{2}-6 y-4 m^{2}-1=0$, thus $m^{2}+1 \equiv 0(\bmod 3)$.
But $m^{2} \equiv 0$ or $1(\bmod 3)$, contradiction.
In summary, the only positive integer solution that satisfies the condition is $(8,7,13)$.