Find all positive integers $n$ for which $n^{n-1} - 1$ is divisible by $2^{2015}$, but not by $2^{2016}$.
To solve the problem, we need to find all positive integers \( n \) such that \( n^{n-1} - 1 \) is divisible by \( 2^{2015} \) but not by \( 2^{2016} \). We will use the Lifting The Exponent Lemma (LTE) to achieve this.

1. **Apply the Lifting The Exponent Lemma (LTE):**
   The LTE lemma states that for odd integers \( x \) and \( y \), and a prime \( p \), the \( p \)-adic valuation of \( x^n - y^n \) is given by:
   \[
   v_p(x^n - y^n) = v_p(x - y) + v_p(n)
   \]
   Here, we need to find \( v_2(n^{n-1} - 1) \). Since \( n \) is a positive integer, we can write:
   \[
   v_2(n^{n-1} - 1) = v_2(n-1) + v_2(n+1) + v_2(n-1) - 1
   \]
   Simplifying, we get:
   \[
   v_2(n^{n-1} - 1) = 2v_2(n-1) + v_2(n+1) - 1
   \]

2. **Set the equation equal to 2015:**
   \[
   2v_2(n-1) + v_2(n+1) - 1 = 2015
   \]
   Adding 1 to both sides, we get:
   \[
   2v_2(n-1) + v_2(n+1) = 2016
   \]

3. **Analyze the gcd condition:**
   Since \( \gcd(n+1, n-1) = 2 \), one of \( n-1 \) or \( n+1 \) must be divisible by \( 2 \) but not both by higher powers of 2. This means:
   - Either \( v_2(n-1) = 1 \)
   - Or \( v_2(n+1) = 1 \)

4. **Case 1: \( v_2(n+1) = 1 \):**
   \[
   2v_2(n-1) + 1 = 2016
   \]
   Subtracting 1 from both sides:
   \[
   2v_2(n-1) = 2015
   \]
   This is impossible because \( 2015 \) is odd and cannot be twice an integer.

5. **Case 2: \( v_2(n-1) = 1 \):**
   \[
   2 \cdot 1 + v_2(n+1) = 2016
   \]
   Simplifying:
   \[
   2 + v_2(n+1) = 2016
   \]
   Subtracting 2 from both sides:
   \[
   v_2(n+1) = 2014
   \]
   This implies that \( n+1 \) is divisible by \( 2^{2014} \) but not by \( 2^{2015} \). Therefore, \( n \) can be written as:
   \[
   n = 2^{2014}k - 1
   \]
   where \( k \) is any odd integer.

Conclusion:
The positive integers \( n \) that satisfy the given conditions are of the form \( n = 2^{2014}k - 1 \) for any odd integer \( k \).

The final answer is \( \boxed{ n = 2^{2014}k - 1 } \) for any odd integer \( k \).