Let $ m,n > 1$ are integers which satisfy $ n|4^m \minus{} 1$ and $ 2^m|n \minus{} 1$. Is it a must that $ n \equal{} 2^{m} \plus{} 1$?
Given the problem, we need to show that if \( m, n > 1 \) are integers satisfying \( n \mid 4^m - 1 \) and \( 2^m \mid n - 1 \), then \( n = 2^m + 1 \).

1. **Express the conditions mathematically:**
   \[
   4^m - 1 = kn \quad \text{for some integer } k
   \]
   \[
   n - 1 = j \cdot 2^m \quad \text{for some integer } j
   \]

2. **Substitute \( n \) from the second condition into the first condition:**
   \[
   n = j \cdot 2^m + 1
   \]
   \[
   4^m - 1 = k(j \cdot 2^m + 1)
   \]

3. **Expand and simplify the equation:**
   \[
   4^m - 1 = kj \cdot 2^m + k
   \]
   \[
   4^m - 1 = k \cdot 2^m \cdot j + k
   \]

4. **Rearrange to isolate \( k \):**
   \[
   4^m - 1 = k(2^m j + 1)
   \]

5. **Consider the modulo \( 2^m \) to find \( k \):**
   \[
   4^m - 1 \equiv 0 \pmod{2^m}
   \]
   \[
   k(2^m j + 1) \equiv 0 \pmod{2^m}
   \]

6. **Since \( 2^m j + 1 \equiv 1 \pmod{2^m} \), we need \( k \equiv 0 \pmod{2^m} \):**
   \[
   k = 2^m p \quad \text{for some integer } p
   \]

7. **Substitute \( k = 2^m p \) back into the equation:**
   \[
   4^m - 1 = (2^m p)(2^m j + 1)
   \]
   \[
   4^m - 1 = 2^m p \cdot 2^m j + 2^m p
   \]
   \[
   4^m - 1 = 2^{2m} p j + 2^m p
   \]

8. **Compare coefficients to solve for \( p \) and \( j \):**
   \[
   4^m - 1 = 2^{2m} p j + 2^m p
   \]
   \[
   4^m - 1 = 2^m (2^m p j + p)
   \]

9. **Since \( 4^m = (2^m)^2 \), we have:**
   \[
   4^m - 1 = 2^m (2^m p j + p)
   \]
   \[
   4^m - 1 = 2^m (p(2^m j + 1))
   \]

10. **For the equation to hold, \( p = 1 \) and \( j = 1 \):**
    \[
    4^m - 1 = 2^m (2^m \cdot 1 + 1)
    \]
    \[
    4^m - 1 = 2^m (2^m + 1)
    \]

11. **Thus, \( n = 2^m + 1 \):**
    \[
    n = 2^m + 1
    \]

The final answer is \( \boxed{ n = 2^m + 1 } \)