$p(2p+1)=(m-3)(m+3)$. Obviosly $p<m<2p+2$. If $m=kp+3,k\ge 3$, then $k=1,2p+1=p+6\to p=5$. If $p=2p-3$, then $2p+1=2(2p-6)$ - contradion (odd=even).

$p(2p+1)=(m+3)(m-3)$ Case1:$p\mid m-3$ Let $m-3=pk,m+3=\frac{2p+1}{k}(k\in \mathbb N)$.If $k\ge 2,$then $m-3\ge 2p\ge p+1\ge \frac{2p+1}{k}=m+3$ which is absurd.Thus $k=1\implies m-3=p,m+3=2p+1\implies p=5\implies m=8$.Therefore $(p,m)=(5,8)$ which satisfies the condition. Case2:$p\mid m+3$ Let $m+3=pk,m-3=\frac{2p+1}{k}$.Obviously $k\ge 2$. $6=pk-\frac{2p+1}{k}\ge pk+(-p-1)=(k-1)p-1$ $\implies 7\ge (k-1)p\implies p\le 7$.Checking $p=2,3,5,7$ cases yields only $p=5$ case satisfies the condition.Therefore $(p,m)=(5,8)$.But $5\nmid 8+3=11$ which is absurd. From case1,2,the answer is $\boxed{(p,m)=(5,8)}\blacksquare$

The solutions are endless. http://www.artofproblemsolving.com/community/c3046h1048216__

Amir Hossein wrote: Find all prime numbers $p$ and positive integers $m$ such that $2p^2 + p + 9 = m^2.$

correct me