Find all positive integers $ m$ with the property that the fourth power of the number of (positive) divisors of $ m$ equals $ m$.
We need to find all positive integers \( m \) such that the fourth power of the number of positive divisors of \( m \) equals \( m \). 

1. Let \( m \) have the prime factorization \( m = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k} \). The number of positive divisors of \( m \) is given by:
   \[
   d(m) = (e_1 + 1)(e_2 + 1) \cdots (e_k + 1)
   \]
   We need:
   \[
   d(m)^4 = m
   \]

2. Substituting \( d(m) \) into the equation, we get:
   \[
   [(e_1 + 1)(e_2 + 1) \cdots (e_k + 1)]^4 = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}
   \]

3. Since \( m \) must be a fourth power, let \( m = n^4 \). Then \( d(m) = n \), and we need:
   \[
   n^4 = m
   \]
   This implies:
   \[
   n = (e_1 + 1)(e_2 + 1) \cdots (e_k + 1)
   \]

4. We need to find \( m \) such that:
   \[
   p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k} = [(e_1 + 1)(e_2 + 1) \cdots (e_k + 1)]^4
   \]

5. Consider the simplest case where \( m \) is a power of a single prime \( p \). Let \( m = p^{4a} \). Then:
   \[
   d(m) = 4a + 1
   \]
   We need:
   \[
   (4a + 1)^4 = p^{4a}
   \]

6. For \( p = 2 \), \( 3 \), or \( 5 \), we check if there are any solutions:
   - For \( p = 2 \):
     \[
     (4a + 1)^4 = 2^{4a}
     \]
     This equation has no integer solutions because \( 4a + 1 \) is odd and \( 2^{4a} \) is even.

   - For \( p = 3 \):
     \[
     (4a + 1)^4 = 3^{4a}
     \]
     We need \( 4a + 1 = 3^k \) for some integer \( k \). Testing small values:
     - \( k = 1 \): \( 4a + 1 = 3 \Rightarrow 4a = 2 \Rightarrow a = \frac{1}{2} \) (not an integer)
     - \( k = 2 \): \( 4a + 1 = 9 \Rightarrow 4a = 8 \Rightarrow a = 2 \)
       \[
       m = 3^{4 \cdot 2} = 3^8 = 6561
       \]

   - For \( p = 5 \):
     \[
     (4a + 1)^4 = 5^{4a}
     \]
     We need \( 4a + 1 = 5^k \) for some integer \( k \). Testing small values:
     - \( k = 1 \): \( 4a + 1 = 5 \Rightarrow 4a = 4 \Rightarrow a = 1 \)
       \[
       m = 5^{4 \cdot 1} = 5^4 = 625
       \]

7. Consider the case where \( m \) is a product of different primes. Let \( m = p_1^{4a_1} p_2^{4a_2} \cdots p_k^{4a_k} \). Then:
   \[
   d(m) = (4a_1 + 1)(4a_2 + 1) \cdots (4a_k + 1)
   \]
   We need:
   \[
   [(4a_1 + 1)(4a_2 + 1) \cdots (4a_k + 1)]^4 = p_1^{4a_1} p_2^{4a_2} \cdots p_k^{4a_k}
   \]

8. Testing combinations of small primes:
   - For \( p_1 = 3 \) and \( p_2 = 5 \):
     \[
     m = 3^{4 \cdot 2} \cdot 5^{4 \cdot 1} = 3^8 \cdot 5^4 = 6561 \cdot 625 = 4100625
     \]

Thus, the solutions are \( m = 625 \), \( 6561 \), and \( 4100625 \).

The final answer is \( \boxed{625, 6561, 4100625} \).