How to find integer solutions of the equation $x(x+9)=y(y+6)$ where x,y are integers? I found that the equation becomes $$(2x+9)^2-(2y+6)^2=45$$. And $45 = 9^2-6^2 = 7^2-{2^2}$. From here I found these solutions $(x,y)=(0,0), \ (0,-6), \ (-9,0),\ (-9,-6),\ (-1,-2),\ (-1, 4), \ (-8,-2), \ (-8,-4)$.
Does this equation has finite numbers of integer solutions and what would be better approach to solve such problems?

The equation in the title (which is different from the one in your text) can be written as
$$(x^2-y^2)+6(x-y)=0 \implies (x-y)(x+y+6)=0.$$
Now either $x=y$ or $x+y+6=0$.
So it has solutions as $\{(a,a) \, | \, a \in \Bbb{Z}\} \cup \{(a,-6-a)\, |\, a \in \Bbb{Z}\}$.
After you modified your original problem
The equation can be written as
$$(2x+2y+15)(2x-2y+3)=45.$$
Now factor $45=ab$ and solve the system
\begin{align*}
2x+2y+15&=a\\
2x-2y+3&=b
\end{align*}
To get
$$x=\frac{a+b-18}{4} \quad \text{ and } \quad y=\frac{a-b-12}{4}.$$
So test the factors $a,b$ of $45$ such that these quantities are integers.
\begin{align*}
ab&=45\\
a+b & \equiv 2 \pmod{4}\\
a-b & \equiv 0 \pmod{4}\\
\end{align*}
Hopefully you can take it from here and see for example $a=9,b=5$ is a solution.
