13. B. Find all prime numbers $p$ and positive integers $m$ that satisfy $2 p^{2}+p+8=m^{2}-2 m$.
13. B. From the given, we have
$$
p(2 p+1)=(m-4)(m+2) \text {. }
$$

Thus, $p \mid(m-4)(m+2)$.
Since $p$ is a prime number, we have
$$
p \mid(m-4) \text { or } p \mid(m+2) \text {. }
$$
(1) If $p \mid(m-4)$, let $m-4=k p$ (where $k$ is a positive integer). Then, $m+2>k p$.
Thus, $3 p^{2}>p(2 p+1)$
$$
=(m-4)(m+2)>k^{2} p^{2} \text {. }
$$

Hence, $k^{2}<3$. Therefore, $k=1$.
Then, $\left\{\begin{array}{l}m-4=p, \\ m+2=2 p+1,\end{array}\right.$ which is impossible.

(2) If $p \mid(m+2)$, let $m+2=k p$ (where $k$ is a positive integer). Then, $m-4<k p$.
Thus, $3 p^{2}>p(2 p+1)$
$$
=(m-4)(m+2)<k^{2} p^{2} \text {. }
$$

Hence, $k^{2}<3$. Therefore, $k=1$.
Then, $\left\{\begin{array}{l}m+2=p, \\ m-4=2 p+1,\end{array}\right.$ which is impossible.

(3) If $p \mid(m-4)$ and $p \mid(m+2)$, then $p \mid 6$.
Thus, $p=2$ or $p=3$.
When $p>5$, we have
$m-4=k p-6>k p-p=p(k-1)$.
Then, $3 p^{2}>p(2 p+1)=(m-4)(m+2)$
$>k(k-1) p^{2}$.
Hence, $k(k-1)<3$. Therefore, $k=1$ or 2.
Since $p(2 p+1)=(m-4)(m+2)$ is odd, we have $k \neq 2$.
Thus, $k=1$.
Then, $\left\{\begin{array}{l}m-4=2 p+1, \\ m+2=p,\end{array}\right.$ which is impossible.

When $p=5$, $m^{2}-2 m=63$, solving gives $m=9$;
When $p=3$, $m^{2}-2 m=29$, no positive integer solution;
When $p=2$, $m^{2}-2 m=18$, no positive integer solution.
In summary, the required prime number $p=5$, and the positive integer $m=9$.