Find all pairs of integers $m, n \geq 2$ such that $$n\mid 1+m^{3^n}+m^{2\cdot 3^n}.$$
1. **Claim**: The solutions are \((m, n) = (3t - 2, 3)\) for any positive integer \(t\). We will verify this claim and show that no other solutions exist.

2. **Verification for \(n = 3\)**:
   - We need to check if \(1 + m^{27} + m^{54}\) is divisible by 3.
   - Note that \(m^2 \equiv 1 \pmod{3}\) for any integer \(m\) not divisible by 3. This is because the only quadratic residues modulo 3 are 0 and 1.
   - Therefore, \(m^{27} \equiv (m^2)^{13} \cdot m \equiv 1^{13} \cdot m \equiv m \pmod{3}\).
   - Similarly, \(m^{54} \equiv (m^2)^{27} \equiv 1^{27} \equiv 1 \pmod{3}\).
   - Thus, \(1 + m^{27} + m^{54} \equiv 1 + m + 1 \equiv m + 2 \pmod{3}\).
   - For \(1 + m^{27} + m^{54}\) to be divisible by 3, we need \(m + 2 \equiv 0 \pmod{3}\), which implies \(m \equiv -2 \equiv 1 \pmod{3}\).

3. **General Case for \(n \geq 5\)**:
   - Let \(p\) be any prime divisor of \(n\). Since \(n \mid (1 + m^{3^n} + m^{2 \cdot 3^n})\), it follows that \(p \mid (m^{3^{n + 1}} - 1)\), so \(m^{3^{n + 1}} \equiv 1 \pmod{p}\).
   - Let \(\operatorname{ord}_m(p) = r\). Then, \(r \mid 3^{n + 1}\). Suppose that \(r = 3^{n + 1}\). By Fermat's Little Theorem, \(m^{p - 1} \equiv 1 \pmod{p}\). Then, \(\operatorname{ord}_m(p) = 3^{n + 1} \mid (p - 1)\). This implies that \(p - 1 \ge 3^{n + 1}\), a contradiction as \(p\) is a divisor of \(n\).
   - Thus, \(r \mid 3^n\). Then, \(m^{3^n} \equiv 1 \pmod{p}\). Multiplying this for every prime factor of \(n\), we get that \(m^{3^n} \equiv 1 \pmod{n}\). Plugging this into the condition \(n \mid (1 + m^{3^n} + m^{2 \cdot 3^n})\), we get that \(n \mid 3\), contradiction.

4. **Conclusion**:
   - The only solutions are when \(n = 3\) and \(m \equiv 1 \pmod{3}\), which can be written as \(m = 3t - 2\) for any positive integer \(t\).

The final answer is \( \boxed{ (m, n) = (3t - 2, 3) } \) for any positive integer \(t\).