Find all pairs of positive integers $(m, n)$ such that $m^2-mn+n^2+1$ divides both numbers $3^{m+n}+(m+n)!$ and $3^{m^3+n^3}+m+n$.

[i]Proposed by Dorlir Ahmeti[/i]
1. Let \( p = m^2 - mn + n^2 + 1 \). We need to show that \( p \) divides both \( 3^{m+n} + (m+n)! \) and \( 3^{m^3+n^3} + m+n \).

2. First, we claim that \( p \) is an odd prime number. Since \( p \) divides \( 3^{m+n} + (m+n)! \), and \( 3^{m+n} + (m+n)! \) is odd for \( m+n \geq 2 \), \( p \) must be odd.

3. Suppose for the sake of contradiction that \( p \) is composite. Then, \( p \) has a prime factor \( q \) such that \( q \leq \sqrt{p} \leq m+n \). This implies \( q \) divides \( (m+n)! \). Since \( q \mid p \mid 3^{m+n} + (m+n)! \), it follows that \( q \mid 3 \), so \( q = 3 \).

4. However, if \( q = 3 \), then \( 3 \) divides \( p = m^2 - mn + n^2 + 1 \). But \( p \equiv m^2 - mn + n^2 + 1 \pmod{3} \). Since \( m \) and \( n \) are integers, \( m^2 \equiv 0, 1 \pmod{3} \) and similarly for \( n^2 \). Therefore, \( m^2 - mn + n^2 + 1 \equiv 0, 1, 2 \pmod{3} \), and it cannot be \( 0 \pmod{3} \) because \( x^2 + 1 \) is never divisible by \( 3 \). This is a contradiction, so \( p \) must be prime.

5. By Fermat's Little Theorem, \( 3^{p-1} \equiv 1 \pmod{p} \). Since \( p = m^2 - mn + n^2 + 1 \), we have \( 3^{m^3 + n^3} \equiv 3^{(m^2 - mn + n^2)(m+n)} \equiv 1 \pmod{p} \). Hence, \( p \) divides \( m+n+1 \).

6. Now, we need \( p = m^2 - mn + n^2 + 1 \leq m+n+1 \). This simplifies to:
   \[
   m^2 - mn + n^2 \leq m + n
   \]
   Rearranging, we get:
   \[
   (m-1)^2 + (n-1)^2 + (m-n)^2 \leq 2
   \]
   The possible integer solutions to this inequality are \( (1,1) \), \( (1,2) \), \( (2,1) \), and \( (2,2) \).

7. Checking these solutions:
   - For \( (1,1) \), \( p = 1^2 - 1 \cdot 1 + 1^2 + 1 = 2 + 1 = 3 \), which is not odd.
   - For \( (1,2) \) and \( (2,1) \), \( p = 1^2 - 1 \cdot 2 + 2^2 + 1 = 1 - 2 + 4 + 1 = 4 \), which is not prime.
   - For \( (2,2) \), \( p = 2^2 - 2 \cdot 2 + 2^2 + 1 = 4 - 4 + 4 + 1 = 5 \), which is prime.

8. Finally, for \( m = n = 2 \):
   \[
   p = 2^2 - 2 \cdot 2 + 2^2 + 1 = 5
   \]
   and it divides \( 3^{m+n} + (m+n)! = 3^4 + 4! = 81 + 24 = 105 \) and \( 3^{m^3 + n^3} + m+n = 3^{8+8} + 4 = 3^{16} + 4 \).

Thus, the only solution is \( (2, 2) \).

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