Solution:

Writing $x^2 = p + q$, $y^2 = p + 7q$, we have $6q = y^2 - x^2 = (y - x)(y + x)$. Since $6q$ is even, one of the factors $y - x$, $y + x$ is even, and then the other is as well; thus $6q$ is divisible by $4 \Rightarrow q$ is even $\Rightarrow q = 2$ and $6q = 12$. We may assume $x, y$ are both taken to be positive; then we must have $y - x = 2$, $y + x = 6 \Rightarrow x = 2$, so $p + 2 = 2^2 = 4 \Rightarrow p = 2$ also.