Find all pairs of prime numbers $(p,q)$ such that for each pair $(p,q)$, there is a positive integer m satisfying
$\frac{pq}{p + q}=\frac{m^2 + 6}{m + 1}$.

To find all pairs of prime numbers \((p, q)\) such that there exists a positive integer \(m\) satisfying the equation 

\[
\frac{pq}{p + q} = \frac{m^2 + 6}{m + 1},
\]

we start by simplifying the given equation. Cross-multiplying gives:

\[
pq(m + 1) = (p + q)(m^2 + 6).
\]

Expanding both sides, we get:

\[
pqm + pq = m^2p + m^2q + 6p + 6q.
\]

Rearranging terms, we obtain:

\[
m^2(p + q) - m(pq) + (pq - 6p - 6q) = 0.
\]

This is a quadratic equation in \(m\):

\[
m^2(p + q) - m(pq) + (pq - 6p - 6q) = 0.
\]

For \(m\) to be an integer, the discriminant of this quadratic equation must be a perfect square. The discriminant \(\Delta\) is given by:

\[
\Delta = (pq)^2 - 4(p + q)(pq - 6p - 6q).
\]

Simplifying the discriminant:

\[
\Delta = p^2q^2 - 4(p + q)(pq - 6p - 6q).
\]

Expanding and simplifying further:

\[
\Delta = p^2q^2 - 4pq(p + q) + 24(p + q)^2.
\]

Let \(a = p + q\) and \(b = pq\). Then the discriminant becomes:

\[
\Delta = b^2 - 4ab + 24a^2.
\]

For \(\Delta\) to be a perfect square, we set:

\[
\Delta = k^2 \quad \text{for some integer } k.
\]

Thus,

\[
b^2 - 4ab + 24a^2 = k^2.
\]

We need to find pairs of primes \((p, q)\) such that this equation holds. Let's test small prime values for \(p\) and \(q\).

1. **Case \(p = 2\):**
   - If \(p = 2\), then \(q\) must be a prime number.
   - Substitute \(p = 2\) into the equation:

\[
\frac{2q}{2 + q} = \frac{m^2 + 6}{m + 1}.
\]

Cross-multiplying:

\[
2q(m + 1) = (2 + q)(m^2 + 6).
\]

Expanding and simplifying:

\[
2qm + 2q = 2m^2 + 12 + qm^2 + 6q.
\]

Rearranging terms:

\[
2m^2 + qm^2 - 2qm + 12 - 4q = 0.
\]

This is a quadratic in \(m\):

\[
m^2(2 + q) - m(2q) + (12 - 4q) = 0.
\]

The discriminant \(\Delta\) is:

\[
\Delta = (2q)^2 - 4(2 + q)(12 - 4q).
\]

Simplifying:

\[
\Delta = 4q^2 - 4(2 + q)(12 - 4q).
\]

Expanding:

\[
\Delta = 4q^2 - 4(24 - 8q + 12q - 4q^2).
\]

Simplifying further:

\[
\Delta = 4q^2 - 4(24 + 4q - 4q^2).
\]

\[
\Delta = 4q^2 - 96 - 16q + 16q^2.
\]

\[
\Delta = 20q^2 - 16q - 96.
\]

For \(\Delta\) to be a perfect square, we need to find \(q\) such that \(20q^2 - 16q - 96\) is a perfect square. Testing small prime values for \(q\):

- If \(q = 3\):

\[
20(3)^2 - 16(3) - 96 = 180 - 48 - 96 = 36.
\]

Since \(36\) is a perfect square (\(6^2\)), \((p, q) = (2, 3)\) is a solution.

2. **Case \(p = 3\):**
   - If \(p = 3\), then \(q\) must be a prime number.
   - Substitute \(p = 3\) into the equation:

\[
\frac{3q}{3 + q} = \frac{m^2 + 6}{m + 1}.
\]

Cross-multiplying:

\[
3q(m + 1) = (3 + q)(m^2 + 6).
\]

Expanding and simplifying:

\[
3qm + 3q = 3m^2 + 18 + qm^2 + 6q.
\]

Rearranging terms:

\[
3m^2 + qm^2 - 3qm + 18 - 3q = 0.
\]

This is a quadratic in \(m\):

\[
m^2(3 + q) - m(3q) + (18 - 3q) = 0.
\]

The discriminant \(\Delta\) is:

\[
\Delta = (3q)^2 - 4(3 + q)(18 - 3q).
\]

Simplifying:

\[
\Delta = 9q^2 - 4(3 + q)(18 - 3q).
\]

Expanding:

\[
\Delta = 9q^2 - 4(54 - 9q + 18q - 3q^2).
\]

Simplifying further:

\[
\Delta = 9q^2 - 4(54 + 9q - 3q^2).
\]

\[
\Delta = 9q^2 - 216 - 36q + 12q^2.
\]

\[
\Delta = 21q^2 - 36q - 216.
\]

For \(\Delta\) to be a perfect square, we need to find \(q\) such that \(21q^2 - 36q - 216\) is a perfect square. Testing small prime values for \(q\):

- If \(q = 2\):

\[
21(2)^2 - 36(2) - 216 = 84 - 72 - 216 = -204.
\]

Since \(-204\) is not a perfect square, \((p, q) = (3, 2)\) is not a solution.

Thus, the only solution is \((p, q) = (2, 3)\).

The final answer is \( \boxed{ (2, 3) } \).