One, (40 points) Find all prime numbers $p, q$ such that $p^{2}-p+1=q^{3}$.


$$
\begin{array}{l}
\text { Given } q^{3}<p^{2} \Rightarrow q<p. \\
\text { Original equation } \Leftrightarrow p(p-1)=(q-1)\left(q^{2}+q+1\right) \\
\Rightarrow p \mid\left(q^{2}+q+1\right). \\
\text { Let } \frac{q^{2}+q+1}{p}=\frac{p-1}{q-1}=k .
\end{array}
$$

Then $p=k(q-1)+1$
$$
\begin{array}{l}
\Rightarrow k(q-1)+1 \leqslant q^{2}+q+1 \\
\Rightarrow k \leqslant \frac{q^{2}+q}{q-1}<q+3 .
\end{array}
$$

From $q^{2}+q+1=p k$, taking both sides modulo $q-1$ gives
$$
\begin{array}{l}
k \equiv 3(\bmod q-1). \\
\text { Also } k=\frac{q^{2}+q+1}{p}<\frac{q^{2}+q+1}{q}=q+1+\frac{1}{q} \\
\Rightarrow k=3 \\
\Rightarrow p=3 q-2,(3 q-2)(3 q-3)+1=q^{3} \\
\Rightarrow(q-7)(q-1)^{2}=0 \\
\Rightarrow(p, q)=(19,7).
\end{array}
$$