Example 3. Given that $x, y, z$ are all integers greater than 1, and $(x, y)=1$, find the integer solution $(x, y)$ for the equation $z=(0 y+3 x)\left(\frac{1}{y}-\frac{1}{x}\right)$.
$$
\begin{aligned}
n_{1}^{9}: z & =(6 y+5 x)\left(\frac{1}{y}-\frac{1}{x}\right) \\
& =(x-y) \cdot \frac{5 x+6 y}{x y} .
\end{aligned}
$$

Since $z$ is an integer, either $x-y$ or $5 x+6 y$ must be divisible by $x y$.

If $x-y$ and $5 x+6 y$ have a common divisor greater than 1, let one of their prime factors be $p$, then
$$
\begin{array}{l}
x-y=p m, \\
x y=p n .
\end{array}
$$

From (1), $x=y+p m$.
Substituting into (2), we get
$(y+p m) y=p n, y^{2}=p(n-m y)$. From the latter equation, we see that $y$ also has the prime factor $p$; and from (3), we know that $x$ also has the prime factor $p$, which contradicts the given condition $(x, y)=1$.
Therefore, $5 x+6 y$ must be divisible by $x y$.
$$
\text { Let } \frac{5 x+6 y}{x y}=k \text { (an integer), }
$$

From the last equation, $y$ must be a divisor of $5 x$. But since $x, y$ are coprime, $y$ must be a divisor of 5. Since $y>1$, we have $y=5$.
$$
\text { Thus } z=(30+5 x)\left(\frac{1}{5}-\frac{1}{x}\right)=1+x=\frac{30}{x}
$$
$>1$ (since $z>1$). Therefore,
$$
x>\sqrt{30}=5.4 \cdots \text {. }
$$

Also, $\frac{30}{x}$ must be an integer, so $x$ must be a divisor of 30. Since $x$ is coprime with 5, we must have $x=6$, thus $z=2$.
Therefore, the integer solution to the original equation is $(x, y, z)=(6, 5, 2)$.