Subtracting the given equations we get $p(p-1) = 2(y-x)(y+x)$.
From this we conclude
$$p \mid y + x,$$
because otherwise $p$ would be a divisor of $y - x$, and $p - 1$ would be a multiple of number $y + x$, which is impossible (we would have $p - 1 \ge y + x > y - x \ge p$ then).
Since $p > y$ (from the second equation) and $y > x$, we have $2p > y + x$, therefore $p = y + x$.
It follows that $p - 1 = 2(y - x)$. By eliminating $y$ we get $p + 1 = 4x$. By plugging that in the first equation we easily get that the only solution is $p = 7$ ($x = 2, y = 5$).