Find all positive integers $m$ and $n$ such that $(2^{2^{n}}+1)(2^{2^{m}}+1) $ is divisible by $m\cdot n $ .
1. **Lemma Proof**: We start by proving the lemma: If for some prime \( p \) and natural \( t \), \( p \mid 2^{2^t} + 1 \), then \( p > t \).

   - Let \( k \) be the minimal integer such that \( p \mid 2^k - 1 \). This implies \( k \mid p-1 \).
   - Since \( k \mid 2^{t+1} \), we can write \( k = 2^r \) for some \( r \leq t+1 \).
   - If \( r \leq t \), then \( 2^r \mid 2^t \), which implies \( p \mid 2^{2^r} - 1 \mid 2^{2^t} - 1 \). This would mean \( p \mid 2^{2^t} - 1 \), contradicting \( p \mid 2^{2^t} + 1 \). Hence, \( r = t+1 \) and \( k = 2^{t+1} \).
   - Therefore, \( p \geq k + 1 = 2^{t+1} + 1 > t \).

2. **Analyzing the Divisibility Condition**: We need to find all positive integers \( m \) and \( n \) such that \( (2^{2^n} + 1)(2^{2^m} + 1) \) is divisible by \( m \cdot n \).

   - Assume \( n \geq m > 1 \). Let \( p \) be a prime divisor of \( m \). Then \( p \leq m \leq n \).
   - From the lemma, if \( p \mid 2^{2^m} + 1 \), then \( p > m \), which is a contradiction since \( p \leq m \). Therefore, \( m = 1 \).

3. **Case \( m = 1 \)**: We need \( n \mid 5(2^{2^n} + 1) \).

   - For \( n = 1 \), \( 5(2^{2^1} + 1) = 5 \cdot 5 = 25 \), and \( 1 \mid 25 \), so \( (m, n) = (1, 1) \) is a solution.
   - For \( n > 1 \), let \( p \) be a prime divisor of \( n \). By the lemma, \( p > n \), which is a contradiction. Hence, \( n = 5 \).

4. **Case \( n = 5 \)**: We need to check if \( 5 \mid 5(2^{2^5} + 1) \).

   - \( 2^{2^5} + 1 = 2^{32} + 1 \equiv 2 \pmod{5} \), so \( 5 \mid 5(2^{32} + 1) \).
   - Therefore, \( (m, n) = (1, 5) \) is a solution.

5. **Case \( n = 1 \)**: We need to check if \( 1 \mid 5(2^{2^1} + 1) \).

   - \( 5(2^{2^1} + 1) = 5 \cdot 5 = 25 \), and \( 1 \mid 25 \), so \( (m, n) = (5, 1) \) is a solution.

The final answer is \(\boxed{(m, n) = (1, 1), (1, 5), (5, 1)}\).