1. Find all prime pairs $(x, y)$ satisfying $x^{y}-y^{x}=$ $x y^{2}-19$.
1. Solution: $\because x^{y}=y^{x}+x y^{2}-19$ and by Fermat's theorem $x^{y} \equiv x(\bmod y)$,
$\therefore x^{y} \equiv x \equiv-19(\bmod y), \therefore y \mid x+19$, similarly $x|y-19, x y|(x+19)(y-19)$, i.e., $x y \mid 19(y$ $-x-19), \therefore x y \mid y-x-19$.
(1) When $x=2$, $2^{y}=3 y^{2}-19$, from $y \mid x+19$ $=21$ we get $y=3,7$, checking shows that $(2,3)$ and $(2,7)$ are both solutions.
(2) When $y=2$, $x^{2}=2^{x}+4 x-19$, and $x \mid(-17), \therefore x=17$, checking shows that $(17,2)$ is not a solution.
(3) When $x \geqslant 3$, when $y \geqslant 3$, from $x y \mid y-x-$ 19 and $y \leqslant x+19$ we get:
$3 x \leqslant x y \leqslant x+19-y \leqslant x+16$, i.e., $3 x \leqslant x+$ $16, \therefore x \leqslant 8, x=3,5,7$. At this time, from $y \mid x+19=22$, 24,26 , we get $y=11,3,13$, which does not satisfy $x \mid y-19$, not a solution.
$\therefore$ All prime pairs $(x, y)$ that satisfy $x^{y}-y^{x}=x y^{2}-19$ are $(2,3)$ and $(2,7)$.