Determine all natural numbers $m$ and $n$  such that

     \[
     n \cdot (n + 1) = 3^m + s(n) + 1182,
     \]

      where $s(n)$ represents the sum of the digits of the natural number $n$.
1. **Consider the given equation:**
   \[
   n \cdot (n + 1) = 3^m + s(n) + 1182
   \]
   where \( s(n) \) represents the sum of the digits of \( n \).

2. **Analyze the equation modulo 9:**
   \[
   n \cdot (n + 1) \equiv 3^m + s(n) + 1182 \pmod{9}
   \]
   Since \( 1182 \equiv 3 \pmod{9} \), the equation becomes:
   \[
   n \cdot (n + 1) \equiv 3^m + s(n) + 3 \pmod{9}
   \]

3. **Possible values of \( 3^m \pmod{9} \):**
   - \( 3^0 \equiv 1 \pmod{9} \)
   - \( 3^1 \equiv 3 \pmod{9} \)
   - \( 3^2 \equiv 0 \pmod{9} \)
   - For \( m \geq 2 \), \( 3^m \equiv 0 \pmod{9} \)

4. **Sum of digits modulo 9:**
   The sum of the digits of \( n \), \( s(n) \), is congruent to \( n \pmod{9} \). Therefore:
   \[
   s(n) \equiv n \pmod{9}
   \]

5. **Substitute \( s(n) \equiv n \) into the equation:**
   \[
   n \cdot (n + 1) \equiv 3^m + n + 3 \pmod{9}
   \]
   Simplifying, we get:
   \[
   n(n + 1) - n - 3 \equiv 3^m \pmod{9}
   \]
   \[
   n^2 - 3 \equiv 3^m \pmod{9}
   \]

6. **Possible values of \( n^2 \pmod{9} \):**
   The possible residues of \( n^2 \pmod{9} \) are \( 0, 1, 4, 7 \).

7. **Check each case for \( 3^m \):**
   - If \( 3^m \equiv 0 \pmod{9} \):
     \[
     n^2 \equiv 3 \pmod{9} \quad \text{(impossible, as 3 is not a quadratic residue modulo 9)}
     \]
   - If \( 3^m \equiv 1 \pmod{9} \):
     \[
     n^2 \equiv 4 \pmod{9} \quad \text{(possible, as 4 is a quadratic residue modulo 9)}
     \]
   - If \( 3^m \equiv 3 \pmod{9} \):
     \[
     n^2 \equiv 6 \pmod{9} \quad \text{(impossible, as 6 is not a quadratic residue modulo 9)}
     \]

8. **Conclusion from modulo analysis:**
   The only possible value for \( 3^m \) is \( 3^0 = 1 \). Therefore, \( m = 0 \).

9. **Substitute \( m = 0 \) into the original equation:**
   \[
   n(n + 1) = s(n) + 1183
   \]

10. **Estimate bounds for \( n \):**
    - Since \( s(n) \leq n \), we have:
      \[
      n(n + 1) \leq n + 1183
      \]
      \[
      n^2 + n - n - 1183 \leq 0
      \]
      \[
      n^2 \leq 1183
      \]
      \[
      n \leq \sqrt{1183} \approx 34.4
      \]
      Thus, \( n \leq 34 \).

    - Also, since \( n(n + 1) > 1183 \):
      \[
      n(n + 1) > 1183
      \]
      \[
      n^2 + n > 1183
      \]
      \[
      n^2 > 1183 - n
      \]
      For \( n \geq 34 \):
      \[
      34(34 + 1) = 34 \cdot 35 = 1190 > 1183
      \]
      Thus, \( n \geq 34 \).

11. **Combine bounds:**
    The only possible value for \( n \) is \( n = 34 \).

12. **Verify the solution:**
    - For \( n = 34 \):
      \[
      34 \cdot 35 = 1190
      \]
      \[
      s(34) = 3 + 4 = 7
      \]
      \[
      3^0 + 7 + 1182 = 1 + 7 + 1182 = 1190
      \]
      The equation holds true.

Therefore, the only solution is \( (m, n) = (0, 34) \).

The final answer is \( \boxed{ (m, n) = (0, 34) } \)