8. 6 .

Let $n^{2}=x+51, m^{2}=x+2019(n, m \in \mathbf{N})$.
Then $m^{2}-n^{2}=(m+n)(m-n)$
$$
=1968=2^{4} \times 3 \times 41.
$$

Since $m+n$ and $m-n$ have the same parity, they are both even, and $m+n>m-n$.

Thus, the possible values of the real number pair $(m+n, m-n)$ are 6 pairs:
$$
\begin{array}{l}
\left(2^{3} \times 3 \times 41,2\right),\left(2^{2} \times 3 \times 41,2^{2}\right), \\
\left(2 \times 3 \times 41,2^{3}\right),\left(2^{3} \times 41,2 \times 3\right), \\
\left(2^{2} \times 41,2^{2} \times 3\right),\left(2 \times 41,2^{3} \times 3\right).
\end{array}
$$

Since the ordered pair $(x, y)$ corresponds one-to-one with $(m+n, m-n)$, the number of positive integer pairs sought is 6 pairs.
8. The number of positive integer pairs $(x, y)$ that satisfy $y=\sqrt{x+51}+\sqrt{x+2019}$ is. $\qquad$