Find all positive integers $ m,n $  that satisfy the equation  \[ 3.2^m +1 = n^2 \]
1. Start with the given equation:
   \[
   3 \cdot 2^m + 1 = n^2
   \]
   Rearrange it to:
   \[
   3 \cdot 2^m = n^2 - 1
   \]
   Notice that \(n^2 - 1\) can be factored as:
   \[
   n^2 - 1 = (n-1)(n+1)
   \]
   Therefore:
   \[
   3 \cdot 2^m = (n-1)(n+1)
   \]

2. Since \(3 \cdot 2^m\) is a product of two consecutive integers, one of them must be divisible by 3. We consider two cases:
   - Case 1: \(n-1\) is divisible by 3.
   - Case 2: \(n+1\) is divisible by 3.

3. **Case 1: \(n-1\) is divisible by 3.**
   Let \(n-1 = 3k\) for some integer \(k\). Then \(n = 3k + 1\). Substitute \(n\) into the equation:
   \[
   3 \cdot 2^m = (3k)(3k + 2)
   \]
   Simplify:
   \[
   3 \cdot 2^m = 9k^2 + 6k
   \]
   Divide both sides by 3:
   \[
   2^m = 3k^2 + 2k
   \]
   Since \(2^m\) is a power of 2, \(3k^2 + 2k\) must also be a power of 2. This implies that \(k\) must be a power of 2. Let \(k = 2^a\):
   \[
   2^m = 3(2^a)^2 + 2(2^a) = 3 \cdot 2^{2a} + 2 \cdot 2^a = 2^a(3 \cdot 2^a + 2)
   \]
   For \(2^m\) to be a power of 2, \(3 \cdot 2^a + 2\) must be a power of 2. We test small values of \(a\):
   - If \(a = 1\):
     \[
     3 \cdot 2^1 + 2 = 6 + 2 = 8 = 2^3
     \]
     Thus, \(m = a + 3 = 1 + 3 = 4\) and \(k = 2^1 = 2\). Therefore, \(n = 3k + 1 = 3 \cdot 2 + 1 = 7\). This gives the solution \((m, n) = (4, 7)\).

4. **Case 2: \(n+1\) is divisible by 3.**
   Let \(n+1 = 3k\) for some integer \(k\). Then \(n = 3k - 1\). Substitute \(n\) into the equation:
   \[
   3 \cdot 2^m = (3k - 2)(3k)
   \]
   Simplify:
   \[
   3 \cdot 2^m = 9k^2 - 6k
   \]
   Divide both sides by 3:
   \[
   2^m = 3k^2 - 2k
   \]
   Since \(2^m\) is a power of 2, \(3k^2 - 2k\) must also be a power of 2. This implies that \(k\) must be a power of 2. Let \(k = 2^a\):
   \[
   2^m = 2^a(3 \cdot 2^a - 2)
   \]
   For \(2^m\) to be a power of 2, \(3 \cdot 2^a - 2\) must be a power of 2. We test small values of \(a\):
   - If \(a = 1\):
     \[
     3 \cdot 2^1 - 2 = 6 - 2 = 4 = 2^2
     \]
     Thus, \(m = a + 2 = 1 + 2 = 3\) and \(k = 2^1 = 2\). Therefore, \(n = 3k - 1 = 3 \cdot 2 - 1 = 5\). This gives the solution \((m, n) = (3, 5)\).

Thus, all positive integral solutions are:
\[
\boxed{(m, n) \in \{(3, 5), (4, 7)\}}
\]