Find all positive integers $n$ satisfying the following.
$$2^n-1 \text{ doesn't have a prime factor larger than } 7$$
1. We need to find all positive integers \( n \) such that \( 2^n - 1 \) does not have a prime factor larger than 7. 

2. Let's start by evaluating \( 2^n - 1 \) for small values of \( n \):
   - For \( n = 1 \): 
     \[
     2^1 - 1 = 1
     \]
     which has no prime factors.
   - For \( n = 2 \):
     \[
     2^2 - 1 = 3
     \]
     which is a prime number and less than 7.
   - For \( n = 3 \):
     \[
     2^3 - 1 = 7
     \]
     which is a prime number and equal to 7.
   - For \( n = 4 \):
     \[
     2^4 - 1 = 15 = 3 \times 5
     \]
     which has prime factors 3 and 5, both less than 7.
   - For \( n = 5 \):
     \[
     2^5 - 1 = 31
     \]
     which is a prime number and greater than 7.

3. From the above calculations, we see that \( n = 1, 2, 3, 4 \) satisfy the condition. 

4. To determine if there are any other values of \( n \) that satisfy the condition, we use Zsigmondy's Theorem. Zsigmondy's Theorem states that for any integer \( a > 1 \) and \( n > 1 \), there exists a prime \( p \) such that \( p \) divides \( a^n - 1 \) and \( p \) does not divide \( a^k - 1 \) for any \( k < n \), with some exceptions.

5. For \( a = 2 \), the exceptions to Zsigmondy's Theorem are:
   - \( n = 1 \)
   - \( n = 6 \) (since \( 2^6 - 1 = 63 = 3^2 \times 7 \))

6. Therefore, we need to check \( n = 6 \):
   \[
   2^6 - 1 = 63 = 3^2 \times 7
   \]
   which has prime factors 3 and 7, both less than or equal to 7.

7. According to Zsigmondy's Theorem, for \( n > 6 \), there will always be a prime factor of \( 2^n - 1 \) that is greater than 7.

8. Thus, the values of \( n \) that satisfy the condition are \( n = 1, 2, 3, 4, \) and \( 6 \).

The final answer is \( \boxed{ n = 1, 2, 3, 4, 6 } \)