7. (21st Balkan Mathematical Olympiad) Given that $x, y$ are prime numbers. Find the solutions of the indeterminate equation $x^{2}-y^{2}=x y^{2}-$ 19.
7. If $x=y$, then there is obviously no solution.

From the given equation, we have
$$
x^{y} \equiv -19 \pmod{y} \text{. }
$$

Since $x$ and $y$ are both primes, and $x \neq y$, we have $(x, y)=1$.
By Fermat's Little Theorem, we have
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$x^{y-1} \equiv 1 \pmod{y}$.
Thus, we have $x + 19 \equiv 0 \pmod{y}$.
Similarly, $19 - y \equiv 0 \pmod{x}$.
Since $x - y + 19 \equiv 0 \pmod{y}$ and $x - y + 19 \equiv 0 \pmod{x}$, we have
$x - y + 19 \equiv 0 \pmod{xy}$.
It is easy to see that $x - y + 19 \neq 0$, so we have
$$
x + y + 19 > |x - y + 19| \geqslant xy \text{. }
$$

This implies $(x-1)(y-1) < 0$, which is a contradiction.
Therefore, $x \leqslant 4$.
It is easy to verify that the solutions to the original indeterminate equation are $(2,3)$ and $(2,7)$.