For a real number $a$, $[a]$ denotes the largest integer not exceeding $a$.
Find all positive real numbers  $x$ satisfying the equation
$$x\cdot [x]+2022=[x^2]$$
1. Let \( [x] = a \) and \( x = a + b \) where \( 0 \leq b < 1 \). This means \( a \) is the integer part of \( x \) and \( b \) is the fractional part of \( x \).

2. Substitute \( x = a + b \) into the given equation:
   \[
   x \cdot [x] + 2022 = [x^2]
   \]
   becomes
   \[
   (a + b) \cdot a + 2022 = [(a + b)^2]
   \]

3. Expand and simplify the left-hand side:
   \[
   a^2 + ab + 2022 = [a^2 + 2ab + b^2]
   \]

4. Since \( [a^2 + 2ab + b^2] \) is the integer part of \( a^2 + 2ab + b^2 \), we can write:
   \[
   a^2 + ab + 2022 = a^2 + [2ab + b^2]
   \]

5. Subtract \( a^2 \) from both sides:
   \[
   ab + 2022 = [2ab + b^2]
   \]

6. Since the right-hand side is an integer, \( ab \) must also be an integer. Let \( ab = k \) where \( k \) is an integer.

7. Now, we have:
   \[
   k + 2022 = [2k + b^2]
   \]

8. Since \( 0 \leq b < 1 \), \( b^2 < 1 \). Therefore, \( [2k + b^2] = 2k \) because \( b^2 \) is a fractional part and does not affect the integer part.

9. Thus, the equation simplifies to:
   \[
   k + 2022 = 2k
   \]

10. Solving for \( k \):
    \[
    2022 = k
    \]

11. Since \( k = ab \) and \( k = 2022 \), we have:
    \[
    ab = 2022
    \]

12. Given \( 0 \leq b < 1 \), we find:
    \[
    b = \frac{2022}{a}
    \]

13. Since \( 0 < b < 1 \), we have:
    \[
    0 < \frac{2022}{a} < 1 \implies a > 2022
    \]

14. Therefore, \( a \) must be an integer greater than or equal to 2023. Hence, \( a \geq 2023 \).

15. The solution for \( x \) is:
    \[
    x = a + \frac{2022}{a}
    \]
    where \( a \geq 2023 \) and \( a \in \mathbb{Z} \).

The final answer is \( \boxed{ x = k + \frac{2022}{k} } \) where \( k \geq 2023 \) and \( k \in \mathbb{Z} \).