Determine all non-constant monic polynomials $f(x)$ with integer coefficients for which there exists a natural number $M$ such that for all $n \geq M$, $f(n)$ divides $f(2^n) - 2^{f(n)}$
    
[i] Proposed by Anant Mudgal [/i]
1. **Initial Setup and Assumptions:**
   We are given a non-constant monic polynomial \( f(x) \) with integer coefficients and a natural number \( M \) such that for all \( n \geq M \), \( f(n) \) divides \( f(2^n) - 2^{f(n)} \). We need to determine all such polynomials.

2. **Divisibility Condition:**
   Let \( p \) be a prime such that \( p \mid f(n) \). Then, for all \( k \geq 0 \),
   \[
   p \mid f(n + pk) \implies p \mid f(2^{n+pk}) - 2^{f(n+pk)}.
   \]
   This implies that \( p \mid f(2^{n+k}) - 2^{f(n+k)} \).

3. **Fixing \( r \) and Choosing \( p \):**
   Fix \( r > N \) and choose \( p \) such that \( p \mid f(2^r) \). Pick a large \( m \) such that \( m \equiv r \pmod{p-1} \). Then,
   \[
   p \mid f(2^m).
   \]
   Since \( p \mid f(2^m) - 2^{f(m)} \), we conclude that \( p = 2 \).

4. **Power of 2 Condition:**
   For each \( r > N \), \( f(2^r) \) must be a power of 2. Suppose \( f(0) \neq 0 \). Then,
   \[
   f(2^r) \equiv f(0) \pmod{2^r}.
   \]
   This implies \( f(2^r) \leq |f(0)| \) for all \( r > N \), which is a contradiction since \( f \) is not constant. Therefore, \( f(0) = 0 \).

5. **Form of \( f(x) \):**
   Write \( f(x) = x^d g(x) \) for some \( d > 0 \) and \( g \in \mathbb{Z}[x] \) with \( g(0) \neq 0 \). Since \( g(2^r) \) is a power of 2 for all large \( r \), and \( g(0) \neq 0 \), \( g \) must be a constant function. Thus, \( f(x) = x^d \).

6. **Checking \( f(x) = x^d \):**
   For all odd primes \( p \), substituting \( f(p) = p^d \) gives:
   \[
   p^d \mid 2^{p^d} - 2^{p^d} \implies p \mid 2^{p^d - d} - 1 \implies p \mid 2^{d-1} - 1.
   \]
   By taking \( p \) large enough, we conclude that \( 2^{d-1} = 1 \), hence \( d = 1 \). Therefore, \( f(x) = x \).

7. **Verification:**
   The polynomial \( f(x) = x \) satisfies the given requirement. For \( f(x) = x \), we have:
   \[
   f(n) = n \quad \text{and} \quad f(2^n) = 2^n.
   \]
   Thus,
   \[
   n \mid 2^n - 2^n,
   \]
   which is true for all \( n \geq M \).

The final answer is \( \boxed{ f(x) = x } \).