Find all pairs of natural numbers $(m, n)$ for which $2^m3^n +1$ is the square of some integer.
To find all pairs of natural numbers \((m, n)\) for which \(2^m3^n + 1\) is the square of some integer, we start by setting up the equation:

\[ 2^m3^n + 1 = x^2 \]

1. **Initial Considerations:**
   - If \(m = 1\), then \(2^m3^n + 1 = 2 \cdot 3^n + 1\). For this to be a perfect square, \(2 \cdot 3^n + 1 \equiv 2 \cdot 3^n + 1 \pmod{4}\). Since \(3 \equiv -1 \pmod{4}\), we have \(2 \cdot (-1)^n + 1 \equiv 2(-1)^n + 1 \pmod{4}\). This can never be a perfect square because it will be either \(3\) or \(1\) modulo \(4\), but not a perfect square.
   - If \(m = 2\), then \(2^m3^n + 1 = 4 \cdot 3^n + 1\). For \(n = 1\), \(4 \cdot 3 + 1 = 13\), which is not a perfect square. For \(n > 1\), \(4 \cdot 3^n + 1\) grows too quickly to be a perfect square.

2. **Case \(m > 2\):**
   - We rewrite the equation as \(2^m3^n = (x-1)(x+1)\). Since \(x-1\) and \(x+1\) are consecutive even numbers, one of them must be a power of 2, and the other must be a power of 3 times 2.
   - Let \(x-1 = 2^a\) and \(x+1 = 2^b3^n\) with \(a < b\). Then:
     \[ 2^b3^n - 2^a = 2 \]
     \[ 2^{b-1}3^n - 2^{a-1} = 1 \]

3. **Solving the Diophantine Equation:**
   - We consider two cases: \(3^n = 2^l + 1\) and \(3^n = 2^l - 1\).

4. **Case \(3^n = 2^l + 1\):**
   - Using \(v_3(2^l + 1) = 1 + v_3(l)\), we find that \(l\) must be divisible by \(3^{n-1}\). However, for \(n > 2\), \(2^{3^{n-1}} > 3^n\), which is not possible. For \(n = 2\), we get \(3^2 = 9 = 2^3 + 1\), so \(l = 3\) and \(m = 5\).
   - For \(n = 1\), we solve \(3 = 2^1 + 1\), so \(l = 1\) and \(m = 3\) or \(m = 4\).

5. **Case \(3^n = 2^l - 1\):**
   - Using \(v_3(2^l - 1) = 1 + v_3(\frac{l}{2})\), we find that \(\frac{l}{2}\) must be divisible by \(3^{n-1}\). However, for \(n > 2\), \(2^{3^{n-1}} - 1 > 3^n\), which is not possible.

Hence, the solutions are:
\[ (m, n) = (3, 1), (4, 1), (5, 2) \]

The final answer is \( \boxed{ (3, 1), (4, 1), (5, 2) } \)