Find all prime numbers $p$ and positive integers $m$ such that $2p^2 + p + 9 = m^2.$
We are given the equation \(2p^2 + p + 9 = m^2\) and need to find all prime numbers \(p\) and positive integers \(m\) that satisfy this equation.

First, we rewrite the equation in a factored form:
\[2p^2 + p + 9 = m^2\]

Rearrange the equation:
\[2p^2 + p + 9 - m^2 = 0\]

We can factor this equation as:
\[p(2p + 1) = (m + 3)(m - 3)\]

This follows from the difference of squares:
\[m^2 - 9 = (m + 3)(m - 3)\]

Now, we consider two cases based on the divisibility of \(p\).

**Case 1:** \(p \mid (m - 3)\)

Let \(m - 3 = pk\) for some integer \(k\). Then:
\[m + 3 = \frac{2p + 1}{k}\]

Since \(k\) is a positive integer, we must have \(k \geq 1\). If \(k \geq 2\), then:
\[m - 3 \geq 2p \geq p + 1 \geq \frac{2p + 1}{k} = m + 3\]

This leads to a contradiction because \(m - 3\) cannot be greater than or equal to \(m + 3\). Therefore, \(k = 1\).

If \(k = 1\), then:
\[m - 3 = p\]
\[m + 3 = 2p + 1\]

Solving these equations, we get:
\[m = p + 3\]
\[p + 3 + 3 = 2p + 1\]
\[p + 6 = 2p + 1\]
\[p = 5\]

Substituting \(p = 5\) back into the equation for \(m\):
\[m = 5 + 3 = 8\]

Thus, \((p, m) = (5, 8)\) satisfies the condition.

**Case 2:** \(p \mid (m + 3)\)

Let \(m + 3 = pk\) for some integer \(k\). Then:
\[m - 3 = \frac{2p + 1}{k}\]

Since \(k\) is a positive integer, we must have \(k \geq 2\). We analyze the inequality:
\[6 = pk - \frac{2p + 1}{k} \geq pk + (-p - 1) = (k - 1)p - 1\]
\[7 \geq (k - 1)p\]
\[p \leq \frac{7}{k - 1}\]

Since \(k \geq 2\), we have \(k - 1 \geq 1\), so:
\[p \leq 7\]

We check the prime numbers \(p = 2, 3, 5, 7\):

- For \(p = 2\):
  \[2(2) + 1 = 5\]
  \[m - 3 = \frac{5}{k}\]
  No integer \(k\) satisfies this.

- For \(p = 3\):
  \[2(3) + 1 = 7\]
  \[m - 3 = \frac{7}{k}\]
  No integer \(k\) satisfies this.

- For \(p = 5\):
  \[2(5) + 1 = 11\]
  \[m - 3 = \frac{11}{k}\]
  No integer \(k\) satisfies this.

- For \(p = 7\):
  \[2(7) + 1 = 15\]
  \[m - 3 = \frac{15}{k}\]
  No integer \(k\) satisfies this.

Thus, no solutions are found in this case.

From both cases, the only solution is:
\[\boxed{(p, m) = (5, 8)}\]