Example 2 Find the positive integer solutions of the equation $x^{3}-y^{3}=z^{2}$. Here $y$ is a prime number, and neither 3 nor $y$ is a divisor of $z$.
Solution: The equation is $(x-y)\left(x^{2}+x y+y^{2}\right)=z^{2}$. 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, $\therefore d=1$ or 3 or $y$, but 3 and $y$ are not divisors of $z$,
$$
\therefore d=1,\left\{\begin{array}{l}
x-y=u^{2}, \\
x^{2}+x y+y^{2}=v^{2} .
\end{array}\right.
$$

Multiplying (2) by 4 gives: $(2 x+y)^{2}+3 y^{2}=4 v^{2}$, which means
( i ) $\left\{\begin{array}{l}2 v-2 x-y=1, \\ 2 v+2 x+y=3 y^{2},\end{array}\right.$
(ii) $\left\{\begin{array}{l}2 v-2 x-y=y, \\ 2 v+2 x+y=3 y,\end{array}\right.$
(iii) $\left\{\begin{array}{l}2 v-2 x-y=3, \\ 2 v+2 x+y=y^{2}\end{array}\right.$
Subtracting the two equations in (1) gives $3 y^{2}=2(2 x+y)+1=$ $4 u^{2}+6 y+1=3 u^{2}+6 y+\left(u^{2}+1\right)$,
$\therefore u^{2}+1 \equiv 0(\bmod 3)$, which contradicts $u^{2} \equiv 0$ or $1(\bmod 3)$!
For (ii), i.e., $\left\{\begin{array}{l}v=x+y, \\ v=y-x,\end{array}\right.$, we get $x=0$, which is a contradiction!
For (iii), subtracting the two equations gives $y^{2}-3=4 u^{2}+6 y$, which means
$$
\begin{array}{l}
(y-3)^{2}-4 u^{2}=12, \\
\therefore\left\{\begin{array}{l}
y-2 u-3=3,2,1, \\
y+2 u-3=4,6,12 .
\end{array}\right.
\end{array}
$$

Only the second case has
a solution $(y, u)=(7,1),(x, z)=(8,13)$.
The solution to the original equation is $(x, y, z)=(8,7,13)$.