Find all triples of natural numbers $(a, b, c)$ for which the number $$2^a + 2^b + 2^c + 3$$ is the square of an integer.
1. **Assume \(a \geq b \geq c\)**:
   We start by assuming \(a \geq b \geq c\). This helps us to systematically explore the possible values of \(a\), \(b\), and \(c\).

2. **Consider the case \(c \geq 2\)**:
   If \(c \geq 2\), then \(2^c \equiv 0 \pmod{4}\). Therefore, the expression \(2^a + 2^b + 2^c + 3\) modulo 4 is:
   \[
   2^a + 2^b + 2^c + 3 \equiv 0 + 0 + 0 + 3 \equiv 3 \pmod{4}
   \]
   Since 3 modulo 4 is not a quadratic residue (i.e., it cannot be a perfect square), \(c\) must be less than 2. Hence, \(c = 1\).

3. **Set \(c = 1\)**:
   With \(c = 1\), the expression simplifies to:
   \[
   2^a + 2^b + 2 + 3 = 2^a + 2^b + 5
   \]
   We need \(2^a + 2^b + 5\) to be a perfect square. Let \(2^a + 2^b + 5 = n^2\) for some integer \(n\).

4. **Consider modulo 8**:
   Since \(a \geq b \geq 1\), we have:
   \[
   2^a + 2^b \equiv 0 \pmod{4}
   \]
   Therefore:
   \[
   2^a + 2^b + 5 \equiv 5 \pmod{8}
   \]
   We need \(5 \pmod{8}\) to be a perfect square modulo 8. The quadratic residues modulo 8 are 0, 1, and 4. Since 5 is not among these, we need to check specific values of \(a\) and \(b\).

5. **Check specific values**:
   - If \(a = b = 1\):
     \[
     2^1 + 2^1 + 5 = 2 + 2 + 5 = 9 = 3^2
     \]
     This is a perfect square. Thus, \((a, b, c) = (1, 1, 1)\) is a solution.

   - If \(b = 2\) and \(a \geq 3\):
     \[
     2^a + 2^2 + 5 = 2^a + 4 + 5 = 2^a + 9
     \]
     We need \(2^a + 9 = n^2\). Rearrange to:
     \[
     n^2 - 2^a = 9
     \]
     This can be factored as:
     \[
     (n - 3)(n + 3) = 2^a
     \]
     Since \(2^a\) is a power of 2, \(n - 3\) and \(n + 3\) must be consecutive powers of 2. The only consecutive powers of 2 are 1 and 2, but this does not work here. The next pair is 4 and 8:
     \[
     n - 3 = 4 \quad \text{and} \quad n + 3 = 8 \quad \Rightarrow \quad n = 5
     \]
     \[
     2^a = 8 \quad \Rightarrow \quad a = 3
     \]
     Thus, \((a, b, c) = (3, 2, 1)\) is another solution.

6. **Verify all solutions**:
   The solutions found are \((a, b, c) = (1, 1, 1)\) and \((3, 2, 1)\). We need to verify if there are any other possible solutions. Given the constraints and the modulo analysis, these are the only solutions.

The final answer is \( \boxed{ (1, 1, 1) } \) and \((3, 2, 1)\).