Let $m \ge 2$ be an integer. Find the smallest integer $n>m$ such that for any partition of the set $\{m,m+1,\cdots,n\}$ into two subsets, at least one subset contains three numbers $a, b, c$ such that $c=a^{b}$.
1. **Define the problem and the set \( S \):**
   We need to find the smallest integer \( n > m \) such that for any partition of the set \( \{m, m+1, \ldots, n\} \) into two subsets, at least one subset contains three numbers \( a, b, c \) such that \( c = a^b \).

2. **Consider the set \( S = \{m, m+1, \ldots, m^{m^{m+2}}\} \):**
   Assume there is a partition of \( S \) into two subsets \( A \) and \( B \) such that no subset contains three numbers \( a, b, c \) for which \( c = a^b \).

3. **Assume \( m \in A \):**
   - Since \( m \in A \), \( m^m \) must be in \( B \) to avoid \( c = a^b \) in \( A \).
   - Then, \((m^m)^{m^m} = m^{m^{m+1}} \in A\) to avoid \( c = a^b \) in \( B \).
   - Next, \(\left(m^{m^{m+1}}\right)^m = m^{m^{m+2}} \in B\) to avoid \( c = a^b \) in \( A \).

4. **Check for contradictions:**
   - Note that \(\left(m^m\right)^{m^{m+1}} = m^{m^{m+2}}\), so \( m^{m^{m+2}} \in B \).
   - But then, \( m, m^{m+1}, m^{m^{m+1}} \in A \), which contradicts our assumption that no subset contains three numbers \( a, b, c \) such that \( c = a^b \).

5. **Consider the set \( \{m, m+1, \ldots, m^{m^{m+2}}\} \) partitioned into \( A \) and \( B \):**
   - Let \( A = \{m, m+1, \ldots, m^m - 1\} \cup \{m^{m^{m+1}}, \ldots, m^{m^{m+2}} - 1\} \).
   - Let \( B = \{m^m, \ldots, m^{m^{m+1}} - 1\} \).

6. **Verify the partition:**
   - For \( a, b \in A \), \( a^b \in A \) if \( a < m^m \) and \( b < m^m \), but \( a^b < m^{m^{m+1}} \).
   - For \( c, d \in B \), \( c^d \ge (m^m)^{m^m} > m^{m^{m+1}} - 1 \), so \( c^d \notin B \).

7. **Conclusion:**
   - The smallest integer \( n \) such that for any partition of the set \( \{m, m+1, \ldots, n\} \) into two subsets, at least one subset contains three numbers \( a, b, c \) such that \( c = a^b \) is \( m^{m^{m+2}} \).

The final answer is \( \boxed{ m^{m^{m+2}} } \).