Find all $(m, n)$ positive integer pairs such that both $\frac{3n^2}{m}$ and $\sqrt{n^2+m}$ are integers.
We need to find all positive integer pairs \((m, n)\) such that both \(\frac{3n^2}{m}\) and \(\sqrt{n^2 + m}\) are integers.

1. **Case 1: \(3 \mid m\)**

   Let \(m = 3x\), where \(x\) is an integer. We can write \(x\) in the form \(x = a^2b\), where \(a\) is an integer and \(b\) is a square-free number.

   Since \(m \mid 3n^2\), we have:
   \[
   3x \mid 3n^2 \implies x \mid n^2
   \]
   This implies:
   \[
   a^2b \mid n^2 \implies a^2 \mid n^2 \implies a \mid n \implies n = ac
   \]
   for some integer \(c\).

   Since \(x \mid n^2\), we have:
   \[
   a^2b \mid a^2c^2 \implies b \mid c^2 \implies b \mid c \implies c = bd
   \]
   for some integer \(d\). Therefore, we can write:
   \[
   n = abd
   \]
   and:
   \[
   m = 3a^2b
   \]

   Now, we need \(\sqrt{n^2 + m}\) to be an integer:
   \[
   n^2 + m = (abd)^2 + 3a^2b = a^2(b^2d^2 + 3b)
   \]
   For \(\sqrt{a^2(b^2d^2 + 3b)}\) to be an integer, \(b^2d^2 + 3b\) must be a perfect square. We analyze the expression:
   \[
   (bd)^2 < b^2d^2 + 3b < (bd + 2)^2
   \]
   This implies:
   \[
   b^2d^2 + 3b = b^2d^2 + 2bd + 1 \implies b = 1 \text{ and } d = 1
   \]
   Therefore:
   \[
   n = a \text{ and } m = 3a^2
   \]

2. **Case 2: \(3 \nmid m\)**

   Let \(m = a^2b\), where \(a\) is an integer and \(b\) is a square-free positive integer.

   Since \(m \mid 3n^2\), we have:
   \[
   a^2b \mid 3n^2
   \]
   This implies:
   \[
   a^2 \mid n^2 \implies a \mid n \implies n = ac
   \]
   for some integer \(c\).

   Since \(m \mid n^2\), we have:
   \[
   a^2b \mid a^2c^2 \implies b \mid c^2 \implies b \mid c \implies c = bd
   \]
   for some integer \(d\). Therefore, we can write:
   \[
   n = abd
   \]
   and:
   \[
   m = a^2b
   \]

   Now, we need \(\sqrt{n^2 + m}\) to be an integer:
   \[
   n^2 + m = (abd)^2 + a^2b = a^2(b^2d^2 + b)
   \]
   For \(\sqrt{a^2(b^2d^2 + b)}\) to be an integer, \(b^2d^2 + b\) must be a perfect square. However:
   \[
   (bd)^2 < b^2d^2 + b < (bd + 1)^2
   \]
   This is a contradiction, as \(b^2d^2 + b\) cannot be a perfect square.

Therefore, the only solution is:
\[
\boxed{n = a, m = 3a^2}
\]
for any \(a \in \mathbb{Z}^+\).