Find all the positive integers $x$ and $y$ that satisfy the equation

$$
x(x-y)=8 y-7
$$
The given equation can be written as:

$$
\begin{aligned}
& x(x-y)=8 y-7 \\
& x^{2}+7=y(x+8)
\end{aligned}
$$

Let \( x+8=m, m \in \mathbb{N} \). Then we have: \( x^{2}+7 \equiv 0 \pmod{m} \), and \( x^{2}+8 x \equiv 0 \pmod{m} \). So we obtain that \( 8 x-7 \equiv 0 \pmod{m} \quad(1) \).

Also we obtain \( 8 x+8^{2}=8(x+8) \equiv 0 \pmod{m} \quad(2) \).

From (1) and (2) we obtain \( (8 x+64)-(8 x-7)=71 \equiv 0 \pmod{m} \), therefore \( m \mid 71 \), since 71 is a prime number, we have:

\( x+8=1 \) or \( x+8=71 \). The only accepted solution is \( x=63 \), and from the initial equation we obtain \( y=56 \).

Therefore the equation has a unique solution, namely \( (x, y)=(63,56) \).

Solution 2:

The given equation is \( x^{2}-x y+7-8 y=0 \).

Discriminant is \( \Delta=y^{2}+32 y-28=(y+16)^{2}-284 \) and must be a perfect square. So \( (y+16)^{2}-284=m^{2} \), and it follows \( (y+16)^{2}-m^{2}=284 \), and after some casework, \( y+16-m=2 \) and \( y+16+m=142 \), hence \( y=56, x=63 \).