Given an even positive integer $m$, find all positive integers $n$ for which there exists a bijection $f:[n]\to [n]$ so that, for all $x,y\in [n]$ for which $n\mid mx-y$, $$(n+1)\mid f(x)^m-f(y).$$

Note: For a positive integer $n$, we let $[n] = \{1,2,\dots, n\}$. 

[i]Proposed by Milan Haiman and Carl Schildkraut[/i]
1. **Understanding the Problem:**
   We need to find all positive integers \( n \) such that there exists a bijection \( f: [n] \to [n] \) satisfying the condition:
   \[
   (n+1) \mid f(x)^m - f(y) \quad \text{for all} \quad x, y \in [n] \quad \text{such that} \quad n \mid mx - y.
   \]
   Here, \( m \) is an even positive integer, and \([n] = \{1, 2, \ldots, n\}\).

2. **Initial Observations:**
   - The condition \( n \mid mx - y \) implies \( y \equiv mx \pmod{n} \).
   - The condition \( (n+1) \mid f(x)^m - f(y) \) must hold for all \( x, y \) satisfying the above congruence.

3. **Case Analysis:**
   - We first consider the case when \( n+1 \) is a prime number \( p \). In this case, we can use properties of primitive roots modulo \( p \).

4. **Construction for \( n+1 \) being prime:**
   - Let \( n+1 = p \) where \( p \) is a prime number.
   - Let \( g \) be a primitive root modulo \( p \). Define the bijection \( f \) by \( f(x) = g^x \pmod{p} \).
   - We need to check if \( (p) \mid f(x)^m - f(y) \) for all \( x, y \) such that \( n \mid mx - y \).

5. **Verification:**
   - Since \( g \) is a primitive root modulo \( p \), \( g \) generates all non-zero residues modulo \( p \).
   - For \( x, y \) such that \( n \mid mx - y \), we have \( y \equiv mx \pmod{n} \).
   - Therefore, \( f(y) = g^y \equiv g^{mx} \pmod{p} \).
   - We need to show \( p \mid g^{mxm} - g^{my} \). Since \( y \equiv mx \pmod{n} \), we have \( g^{my} \equiv g^{m(mx)} \equiv (g^m)^{mx} \pmod{p} \).
   - Thus, \( g^{mxm} \equiv g^{my} \pmod{p} \), and the condition holds.

6. **Proof for \( n+1 \) not being prime:**
   - Assume \( n+1 \) is not prime. Then \( n+1 \) can be written as a product of distinct primes \( p_1, p_2, \ldots, p_k \).
   - We will show that this leads to a contradiction.

7. **Contradiction:**
   - Consider an element \( q \in [n] \) such that \( p_1 p_2 \ldots p_{k-1} \mid q \) but \( q \equiv 1 \pmod{p_k} \).
   - There exists \( a \in [n] \) such that \( a^m \equiv q \pmod{n+1} \).
   - By the lemma, there are \( \gcd(m, n) \) such \( a \).
   - Similarly, there are \( \gcd(m, n) \) elements \( y \in [n] \) such that \( y^m \equiv 1 \pmod{n+1} \).
   - Since \( m \) is even, there are more such \( y \) than such \( x \), leading to a contradiction.

8. **Conclusion:**
   - The only possible values for \( n \) are those for which \( n+1 \) is a prime number.

The final answer is \( \boxed{ n = p - 1 } \) where \( p \) is a prime number.