Find all quadruplets of positive integers $(m,n,k,l)$ such that $3^m = 2^k + 7^n$ and $m^k = 1 + k + k^2 + k^3 + \ldots + k^l$.
1. **Analyzing the equation \(3^m = 2^k + 7^n\) for \(k = 1\):**
   - For \(k = 1\), the equation becomes \(3^m = 2 + 7^n\).
   - We need to find positive integers \(m\) and \(n\) such that this equation holds.

2. **Checking small values of \(m\):**
   - For \(m = 1\):
     \[
     3^1 = 3 \quad \text{and} \quad 2 + 7^n = 3 \implies 7^n = 1 \implies n = 0 \quad (\text{not a positive integer})
     \]
     No solution for \(m = 1\).
   - For \(m = 2\):
     \[
     3^2 = 9 \quad \text{and} \quad 2 + 7^n = 9 \implies 7^n = 7 \implies n = 1
     \]
     This gives the solution \((m, n, k) = (2, 1, 1)\).

3. **Considering \(m \geq 3\):**
   - Taking modulo 27:
     \[
     3^m \equiv 2 + 7^n \pmod{27}
     \]
     Since \(3^3 = 27 \equiv 0 \pmod{27}\), for \(m \geq 3\), \(3^m \equiv 0 \pmod{27}\).
     \[
     7^n + 2 \equiv 0 \pmod{27} \implies 7^n \equiv -2 \equiv 25 \pmod{27}
     \]
     Checking powers of 7 modulo 27:
     \[
     7^1 \equiv 7, \quad 7^2 \equiv 22, \quad 7^3 \equiv 19, \quad 7^4 \equiv 25, \quad \ldots
     \]
     So, \(n \equiv 4 \pmod{9}\).

4. **Taking modulo 37:**
   - We need \(3^m \equiv 7^n + 2 \equiv 35 \pmod{37}\).
   - Checking powers of 3 modulo 37:
     \[
     3^1 \equiv 3, \quad 3^2 \equiv 9, \quad 3^3 \equiv 27, \quad 3^4 \equiv 7, \quad \ldots
     \]
     None of these are congruent to 35 modulo 37, so no solution for \(m \geq 3\).

5. **Considering \(k \geq 2\):**
   - Taking modulo 4:
     \[
     3^m \equiv 2^k + 7^n \pmod{4}
     \]
     Since \(3 \equiv -1 \pmod{4}\) and \(7 \equiv -1 \pmod{4}\), we have:
     \[
     (-1)^m \equiv 2^k + (-1)^n \pmod{4}
     \]
     This implies \(m\) and \(n\) must have the same parity.

6. **Taking modulo 7:**
   - We need \(3^m \equiv 2^k \pmod{7}\).
   - Checking powers of 3 modulo 7:
     \[
     3^1 \equiv 3, \quad 3^2 \equiv 2, \quad 3^3 \equiv 6, \quad 3^4 \equiv 4, \quad 3^5 \equiv 5, \quad 3^6 \equiv 1
     \]
     This implies \(m\) must be even.

7. **Expressing \(m\) and \(n\) as even integers:**
   - Let \(m = 2x\) and \(n = 2y\).
   - The equation becomes:
     \[
     (3^x - 7^y)(3^x + 7^y) = 2^k
     \]
     Since \(\gcd(3^x - 7^y, 3^x + 7^y) = 2\), we have:
     \[
     3^x - 7^y = 2 \quad \text{and} \quad 3^x + 7^y = 2^{k-1}
     \]
     Solving these, the only solution is \((x, y) = (2, 1)\), giving \(m = 4\), \(k = 5\), and \(n = 2\).

8. **Checking the second equation \(m^k = 1 + k + k^2 + \ldots + k^l\):**
   - For \((m, n, k) = (2, 1, 1)\), we need to check if there exists \(l\) such that:
     \[
     2^1 = 1 + 1 + 1^2 + \ldots + 1^l
     \]
     This simplifies to \(2 = l\), so \(l = 1\).

Conclusion:
The only solution is \((m, n, k, l) = (2, 1, 1, 1)\).

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