Let $N$ is the set of all positive integers. Determine all mappings $f: N-\{1\} \to N$ such that for every $n \ne m$ the following equation is true $$f(n)f(m)=f\left((nm)^{2021}\right)$$

1. **Restate the problem and introduce notation:**
   We are given a function \( f: \mathbb{N} - \{1\} \to \mathbb{N} \) such that for every \( n \neq m \), the following equation holds:
   \[
   f(n)f(m) = f\left((nm)^{2021}\right)
   \]
   For simplicity, let \( a = 2021 \). Define \( P(x, y) \) as the assertion \( f(x)f(y) = f((xy)^a) \) for all \( x, y \geq 2 \) and \( x \neq y \).

2. **Compare \( P(x^{k+1}, x) \) with \( P(x^k, x^2) \):**
   \[
   P(x^{k+1}, x): f(x^{k+1})f(x) = f((x^{k+1} \cdot x)^a) = f(x^{(k+2)a})
   \]
   \[
   P(x^k, x^2): f(x^k)f(x^2) = f((x^k \cdot x^2)^a) = f(x^{(k+2)a})
   \]
   Since both expressions equal \( f(x^{(k+2)a}) \), we have:
   \[
   f(x^{k+1})f(x) = f(x^k)f(x^2)
   \]

3. **Solve for \( f(x^{k+1}) \):**
   \[
   f(x^{k+1}) = f(x^k) \frac{f(x^2)}{f(x)}
   \]
   This recurrence relation holds for all \( x \geq 2 \) and \( k \geq 3 \).

4. **Express \( f(x^k) \) in terms of \( f(x^3) \):**
   By iterating the recurrence relation, we get:
   \[
   f(x^k) = f(x^3) \left( \frac{f(x^2)}{f(x)} \right)^{k-3}
   \]
   This holds for all \( x \geq 2 \) and \( k \geq 3 \).

5. **Rewrite \( P(x, y) \) using the new expression:**
   \[
   P(x, y): f(x)f(y) = f(x^3 y^3) \left( \frac{f(x^2 y^2)}{f(xy)} \right)^{a-3}
   \]
   This holds for all \( x, y \geq 2 \) and \( x \neq y \).

6. **Consider \( Q(x^u, x^v) \) for \( u, v \geq 3 \) and \( u \neq v \):**
   \[
   Q(x^u, x^v): f(x^u)f(x^v) = f(x^{3u+3v}) \left( \frac{f(x^{2u+2v})}{f(x^{u+v})} \right)^{a-3}
   \]
   Using the expression for \( f(x^k) \):
   \[
   f(x^u) = f(x^3) \left( \frac{f(x^2)}{f(x)} \right)^{u-3}
   \]
   \[
   f(x^v) = f(x^3) \left( \frac{f(x^2)}{f(x)} \right)^{v-3}
   \]
   \[
   f(x^{3u+3v}) = f(x^3) \left( \frac{f(x^2)}{f(x)} \right)^{3u+3v-3}
   \]
   \[
   f(x^{2u+2v}) = f(x^3) \left( \frac{f(x^2)}{f(x)} \right)^{2u+2v-3}
   \]
   \[
   f(x^{u+v}) = f(x^3) \left( \frac{f(x^2)}{f(x)} \right)^{u+v-3}
   \]

7. **Substitute these into \( Q(x^u, x^v) \):**
   \[
   f(x^3) \left( \frac{f(x^2)}{f(x)} \right)^{u-3} f(x^3) \left( \frac{f(x^2)}{f(x)} \right)^{v-3} = f(x^3) \left( \frac{f(x^2)}{f(x)} \right)^{3u+3v-3} \left( \frac{f(x^3) \left( \frac{f(x^2)}{f(x)} \right)^{2u+2v-3}}{f(x^3) \left( \frac{f(x^2)}{f(x)} \right)^{u+v-3}} \right)^{a-3}
   \]
   Simplifying, we get:
   \[
   f(x^3) = \left( \frac{f(x^2)}{f(x)} \right)^{(u+v)(a-1)+3}
   \]

8. **Conclude that \( f(x^2) = f(x) \) and \( f(x^3) = 1 \):**
   Since the above equation must hold for all distinct \( u, v \geq 3 \), we deduce:
   \[
   f(x^2) = f(x) \quad \text{and} \quad f(x^3) = 1 \quad \forall x > 1
   \]
   Therefore, \( f(x^k) = 1 \) for all \( k \geq 3 \).

9. **Simplify \( P(x, y) \) to find \( f(x) \):**
   \[
   P(x, y): f(x)f(y) = 1 \quad \forall \text{ distinct } x, y \geq 2
   \]
   This implies:
   \[
   f(x) = 1 \quad \forall x \geq 2
   \]

Conclusion:
\[
\boxed{f(x) = 1 \quad \forall x \geq 2}
\]