Writing $x^{2}=p+q, y^{2}=p+7 q$, we have $6 q=y^{2}-x^{2}=(y-x)(y+x)$. Since $6 q$ is even, one of the factors $y-x, y+x$ is even, and then the other is as well; thus $6 q$ is divisible by $4 \Rightarrow q$ is even $\Rightarrow q=2$ and $6 q=12$. 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.