Find all integers $n\ge1$ such that $2^n-1$ has exactly $n$ positive integer divisors. 

[i]Proposed by Ankan Bhattacharya [/i]
1. **Initial Observations and Definitions:**
   - For each \( k \in \mathbb{N} \), let \( d(k) \) be the number of divisors, \( \omega(k) \) be the number of distinct prime divisors, and \( \Omega(k) \) be the number of prime divisors counted with multiplicity.
   - We need to find all integers \( n \geq 1 \) such that \( 2^n - 1 \) has exactly \( n \) positive integer divisors.

2. **Parity Consideration:**
   - If \( n \) is odd, then \( d(2^n - 1) \) is odd, implying \( 2^n - 1 \) is a perfect square.
   - However, \( 2^n - 1 \equiv 3 \pmod{4} \) for \( n \geq 2 \), which cannot be a perfect square. Thus, \( n \) must be even unless \( n = 1 \).

3. **Bounding the Number of Distinct Prime Divisors:**
   - Using Zsigmondy's theorem (Bang's theorem), for \( n \neq 1, 6 \), there exists a prime \( p \) such that \( p \mid 2^n - 1 \) and \( p \nmid 2^k - 1 \) for all \( 1 \leq k < n \).
   - For all \( k \mid n \) (except possibly \( k = 1, 6 \)), \( 2^k - 1 \) has a primitive prime divisor. Thus, \( \omega(2^n - 1) \geq d(n) - 2 \).

4. **Prime Factorization and Divisor Counting:**
   - Let \( 2^n - 1 = p_1^{\alpha_1} \cdot p_2^{\alpha_2} \cdots p_k^{\alpha_k} \), where \( p_i \) are distinct primes.
   - The number of divisors of \( 2^n - 1 \) is \( (\alpha_1 + 1)(\alpha_2 + 1) \cdots (\alpha_k + 1) = n \).

5. **Bounding the Number of Prime Divisors Counted with Multiplicity:**
   - From the previous step, \( \Omega(n) \geq d(n) - 2 \).
   - Let \( n = q_1^{\beta_1} q_2^{\beta_2} \cdots q_m^{\beta_m} \). Then, \( \sum_{i=1}^m \beta_i = \Omega(n) \geq d(n) - 2 = (\beta_1 + 1)(\beta_2 + 1) \cdots (\beta_m + 1) - 2 \).

6. **Bounding the Number of Distinct Prime Divisors of \( n \):**
   - If \( n \) has at least 3 distinct prime divisors, then \( \prod_{i=1}^m (\beta_i + 1) \geq \beta_1 \beta_2 + \beta_1 \beta_3 + \beta_2 \beta_3 + \sum_{i=1}^m \beta_i + 1 \), which contradicts \( \sum_{i=1}^m \beta_i \geq (\beta_1 + 1)(\beta_2 + 1) \cdots (\beta_m + 1) - 2 \). Thus, \( \omega(n) \leq 2 \).

7. **Case Analysis for \( \omega(n) = 2 \):**
   - If \( n \) has exactly 2 distinct prime divisors, then \( n = 2^a \cdot q^b \) for some prime \( q \).
   - If \( 3 \nmid n \), then \( \Omega(n) \geq d(n) - 1 \), leading to a contradiction unless \( n = 6 \).

8. **Powers of Two:**
   - If \( n = 2^T \), then \( 2^n - 1 = 2^{2^T} - 1 \) has \( 2^T \) divisors.
   - Fermat numbers \( F_k = 2^{2^k} + 1 \) are prime for \( k = 0, 1, 2, 3, 4 \), but \( F_5 \) is not prime.
   - Thus, \( n = 2, 4, 8, 16, 32 \) work.

9. **Conclusion:**
   - The integers \( n \geq 1 \) such that \( 2^n - 1 \) has exactly \( n \) positive integer divisors are \( n = 1, 2, 4, 6, 8, 16, 32 \).

The final answer is \( \boxed{ n = 1, 2, 4, 6, 8, 16, 32 } \).