8. The number of positive integer pairs $(x, y)$ that satisfy $y=\sqrt{x+51}+\sqrt{x+2019}$ is $\qquad$ pairs.
Let $m^{2}=x+51, n^{2}=x+2019$, then $n^{2}-m^{2}=(n-m)(n+m)=1968=2^{4} \times 3 \times 41$. Since $n-m, n+m$ have the same parity, and $(n-m)(n+m)$ is even, it follows that $n-m, n+m$ are both even. Thus, the combinations of $(n-m, n+m)$ are $\left(2,2^{3} \times 3 \times 41\right),\left(2^{2}, 2^{4} \times 3 \times 41\right),\left(2^{3}, 2 \times 3 \times 41\right),\left(2 \times 3,2^{3} \times\right.$ 41), $\left(2^{2} \times 3,2^{2} \times 41\right),\left(2^{3} \times 3,2 \times 41\right)$, a total of 6 pairs.

For each pair $(n-m, n+m)$, there is a unique pair $(m, n)$ such that $m^{2}=x+51, n^{2}=x+2019$ holds, thus uniquely corresponding to a pair $(x, y)$. Therefore, there are 6 pairs of positive integers $(x, y)$.