Find all monic nonconstant polynomials $P$ with integer coefficients for which there exist positive integers $a$ and $m$ such that for all positive integers $n\equiv a\pmod m$, $P(n)$ is nonzero and $$2022\cdot\frac{(n+1)^{n+1} - n^n}{P(n)}$$ is an integer. 

[i]Jaedon Whyte, Luke Robitaille, and Pitchayut Saengrungkongka[/i]
1. **Identify the given conditions and the goal:**
   We need to find all monic nonconstant polynomials \( P \) with integer coefficients such that there exist positive integers \( a \) and \( m \) for which, for all positive integers \( n \equiv a \pmod{m} \), the expression 
   \[
   2022 \cdot \frac{(n+1)^{n+1} - n^n}{P(n)}
   \]
   is an integer.

2. **Check the given solution for specific polynomials:**
   The solution suggests that \( P(x) = x^2 + x + 1 \) and \( P(x) = (x^2 + x + 1)^2 \) work. We need to verify this.

3. **Verification for \( P(x) = x^2 + x + 1 \):**
   - Take \( a = 1 \) and \( m = 6 \).
   - We need to check if \( x^2 + x + 1 \mid (x+1)^{x+1} - x^x \) for \( x \equiv 1 \pmod{6} \).
   - For \( x \equiv 1 \pmod{6} \), let \( x = 6k + 1 \). Then,
     \[
     (6k+2)^{6k+2} - (6k+1)^{6k+1}
     \]
     We need to show that \( x^2 + x + 1 = (6k+1)^2 + (6k+1) + 1 = 36k^2 + 12k + 3 \) divides the above expression.

4. **Verification for \( P(x) = (x^2 + x + 1)^2 \):**
   - We need to show that if \( p^k \mid x^2 + x + 1 \), then \( p^{2k} \mid (x+1)^{x+1} - x^x \) for \( x \equiv 1 \pmod{6} \).
   - Let \( y = x + 6tp^k \) such that \( p^{2k} \mid y^2 + y + 1 \). Clearly, \( p^{2k} \mid (y+1)^{y+1} - y^y \).
   - We need to check:
     \[
     (x+1)^{x+1} - x^x \equiv (y+1-6tp^k)^{y+1-6tp^k} - (y-6tp^k)^{y-6tp^k} \pmod{p^{2k}}
     \]
     Using lifting the exponent, we can show that:
     \[
     \nu_p((y+1)^{6tp^k} - y^{6tp^k}) \ge k + \nu_p((y+1)^6 - y^6) \ge k + \nu_p(y^2 + y + 1) \ge 2k
     \]
     Thus, \( (y+1)^{y+1-6tp^k} \equiv y^{y-6tp^k} \pmod{p^{2k}} \).

5. **Verification for \( P(x) = (x^2 + x + 1)^3 \):**
   - Pick a large prime \( p > m \) such that \( p \mid x^2 + x + 1 \).
   - We need to show that \( p^3 \mid (x+1)^{x+1} - x^x \).
   - Consider all \( A \) such that \( p^3 \mid (x + Ap + 1)^{x + Ap + 1} - (x + Ap)^{x + Ap} \).
   - Expand and simplify modulo \( p^3 \):
     \[
     \left( \frac{x+1+Ap}{x+1} \right)^{x+1} (x+1+Ap)^{Ap} \equiv \left( \frac{x+Ap}{x} \right)^x (x+Ap)^{Ap} \pmod{p^3}
     \]
     This leads to a contradiction, showing that \( P(x) = (x^2 + x + 1)^3 \) fails.

Conclusion:
The only monic nonconstant polynomials \( P \) with integer coefficients that satisfy the given conditions are \( P(x) = x^2 + x + 1 \) and \( P(x) = (x^2 + x + 1)^2 \).

The final answer is \( \boxed{ P(x) = x^2 + x + 1 } \) and \( P(x) = (x^2 + x + 1)^2 \).