Find all pairs of natural numbers $(m,n)$ bigger than $1$ for which $2^m+3^n$ is the square of whole number.

[i]I. Tonov[/i]
1. **Modulo 3 Analysis:**
   - Consider the equation \(2^m + 3^n = k^2\) where \(k\) is a whole number.
   - Taking modulo 3, we have:
     \[
     2^m + 3^n \equiv k^2 \pmod{3}
     \]
   - Since \(3^n \equiv 0 \pmod{3}\), we get:
     \[
     2^m \equiv k^2 \pmod{3}
     \]
   - The possible values of \(2^m \pmod{3}\) are:
     - If \(m\) is even, \(2^m \equiv 1 \pmod{3}\).
     - If \(m\) is odd, \(2^m \equiv 2 \pmod{3}\).
   - Since \(k^2 \equiv 0, 1 \pmod{3}\), \(2^m \equiv k^2 \pmod{3}\) implies \(m\) must be even.

2. **Modulo 4 Analysis:**
   - Taking modulo 4, we have:
     \[
     2^m + 3^n \equiv k^2 \pmod{4}
     \]
   - Since \(2^m \equiv 0 \pmod{4}\) for \(m \geq 2\), we get:
     \[
     3^n \equiv k^2 \pmod{4}
     \]
   - The possible values of \(3^n \pmod{4}\) are:
     - If \(n\) is even, \(3^n \equiv 1 \pmod{4}\).
     - If \(n\) is odd, \(3^n \equiv 3 \pmod{4}\).
   - Since \(k^2 \equiv 0, 1 \pmod{4}\), \(3^n \equiv k^2 \pmod{4}\) implies \(n\) must be even.

3. **Expressing the Equation:**
   - Let \(m = 2a\) and \(n = 2b\) for some natural numbers \(a\) and \(b\).
   - The equation becomes:
     \[
     2^{2a} + 3^{2b} = k^2
     \]
   - This can be rewritten as:
     \[
     (2^a)^2 + (3^b)^2 = k^2
     \]
   - Let \(X = 2^a\) and \(Y = 3^b\), then:
     \[
     X^2 + Y^2 = k^2
     \]
   - This implies \(X^2 + Y^2\) is a perfect square.

4. **Analyzing the Equation:**
   - Consider the equation \(X^2 + Y^2 = k^2\).
   - This can be factored as:
     \[
     (k - X)(k + X) = Y^2
     \]
   - Let \(k - X = d\) and \(k + X = c\), then \(dc = Y^2\) and \(c - d = 2X\).
   - Since \(Y = 3^b\), we have:
     \[
     d = 1 \quad \text{and} \quad c = Y^2
     \]
   - This implies:
     \[
     k - X = 1 \quad \text{and} \quad k + X = Y^2
     \]
   - Solving these, we get:
     \[
     2k = 1 + Y^2 \quad \text{and} \quad 2X = Y^2 - 1
     \]
   - Since \(Y = 3^b\), we have:
     \[
     2k = 1 + 9^b \quad \text{and} \quad 2X = 9^b - 1
     \]
   - This implies:
     \[
     k = \frac{1 + 9^b}{2} \quad \text{and} \quad X = \frac{9^b - 1}{2}
     \]
   - For \(k\) and \(X\) to be integers, \(1 + 9^b\) and \(9^b - 1\) must be even, which is true for all \(b \geq 1\).

5. **Conclusion:**
   - Since \(m = 2a\) and \(n = 2b\), and \(a, b \geq 1\), the pairs \((m, n)\) are of the form \((2a, 2b)\) where \(a, b \geq 1\).

The final answer is \( \boxed{ (m, n) = (2a, 2b) } \) where \(a, b \geq 1\).