7. The positive integer solutions of the equation $x+y^{2}+(x, y)^{3}=x y(x, y)$ are $\qquad$ groups $((x, y)$ represents the greatest common divisor of integers $x, y)$.


7.4.

Let $(x, y)=d$. Then $d^{2} \mid x$.
Let $x=a d^{2}, y=b d$. Then $(a d, b)=1$.
Thus, the original equation becomes
$$
a+b^{2}+d=a b d^{2} \text {. }
$$

Therefore, $b \mid(a+d) \Rightarrow b \leqslant a+d$.
Then $a+b^{2}+d=a b d^{2}$
$$
\begin{array}{l}
=(a+d) b+(a+d) b+b\left(a d^{2}-2 a-2 d\right) \\
\geqslant a+d+b^{2}+b\left(a d^{2}-2 a-2 d\right) .
\end{array}
$$

Thus, $a d^{2}-2 a-2 d \leqslant 0$.
If $d \geqslant 3$, then
$$
\begin{aligned}
a d^{2}- & 2 a-2 d \geqslant 3 a d-2 a-2 d \\
& =2 a d+a d-2 a-2 d \\
& \geqslant 2 d+3 a-2 a-2 d>0,
\end{aligned}
$$

which is a contradiction.
Therefore, $d \leqslant 2$.
(1) $d=2$.
From $a d^{2}-2 a-2 d \leqslant 0$, we get $a \leqslant 2$.
When $a=2$, from equation (1) we get $b^{2}-8 b+4=0$, in which case, $b$ has no integer solutions;
When $a=1$, from equation (1) we get
$$
b^{2}-4 b+3=0 \Rightarrow b=1 \text { or } 3 \text {, }
$$

In this case, the solutions to the original equation are
$$
(x, y)=(4,2) \text { or }(4,6) \text {. }
$$
(2) $d=1$.

From equation (1) we get
$$
\begin{array}{l}
2=(a-b-1)(b-1) \\
\Rightarrow(a-b-1, b-1)=(2,1) \text { or }(1,2) \\
\Rightarrow(a, b)=(5,2) \text { or }(5,3),
\end{array}
$$

In this case, the solutions to the original equation are
$$
(x, y)=(5,2) \text { or }(5,3) \text {. }
$$