Example 9 (15th All-Soviet Union Mathematical Olympiad) Find all positive integer solutions to the equation $x^{3}-y^{3}=x y+61$.

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Solution: Clearly, $x>y$. Let $x=y+d$ ($d$ is a positive integer), then
$$
\begin{array}{l}
(y+d)^{3}-y^{3}=(y+d) y+61, \\
(3 d-1) y^{2}+d(3 d-1) y+d^{3}=61 .
\end{array}
$$

From (1), we know $d^{3}<61$, hence, $d \leqslant 3$.
If $d=1$, then (1) becomes $2 y^{2}+2 y-60=0, y^{2}+y-30=0$, solving this gives $y=5, y=-6$ (discard), thus $x=6$.
If $d=2$, then (1) becomes $5 y^{2}+10 y-53=0$, this equation has no positive integer solutions.
If $d=3$, then (1) becomes $8 y^{2}+24 y-34=0$, this equation also has no positive integer solutions.

In conclusion, the original equation has only one set of positive integer solutions $(x, y)=(6,5)$.