8. Let $p, q$ be prime numbers, and satisfy $p^{3}+q^{3}+1=p^{2} q^{2}$. Then the maximum value of $p+q$ is
8.5.

Assume $q \leqslant p$. Notice that
$$
\begin{array}{l}
p^{3}+q^{3}+1=p^{2} q^{2} \\
\Rightarrow q^{3}+1=p^{2} q^{2}-p^{3} \\
\Rightarrow(q+1)\left(q^{2}-q+1\right)=p^{2}\left(q^{2}-p\right) .
\end{array}
$$

Therefore, $p^{2} \mid(q+1)\left(q^{2}-q+1\right)$.
Since $q \leqslant p$, we have $0<q^{2}-q+1<p^{2}$.
This implies $p \mid(q+1)$.
Also, since $q \leqslant p$, we have $q+1=p$.
Given that $p$ and $q$ are both prime numbers, the only solution is $p=3$ and $q=2$. Thus, the maximum value of $p+q$ is 5.