4-167 Find the positive integer solutions of the equation $x^{3}-y^{3}=x y+61$.

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［Solution］Obviously $x>y$. Let $x=y+d(d \geqslant 1)$.
Substitute into the original equation to get
$$
\begin{array}{l}
(y+d)^{3}-y^{3}=(y+d) y+61 \\
(3 d-1) y^{2}+\left(3 d^{2}-d\right) y+d^{3}=61 .
\end{array}
$$

From (1), we get $d^{3}<61$,
thus $\quad d \leqslant 3$.
If $d=1$, then (1) becomes
$$
\begin{array}{l}
2 y^{2}+2 y-60=0, \\
y^{2}+y-30=0, \\
(y+6)(y-5)=0,
\end{array}
$$

Thus, we get $y=5, x=6$.
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 summary, the original equation has only one set of positive integer solutions.
$$
\left\{\begin{array}{l}
x=6 \\
y=5 .
\end{array}\right.
$$