A positive integer $m$ is said to be $\emph{zero taker}$ if there exists a positive integer $k$ such that:

$k$ is a perfect square;
$m$ divides $k$;
the decimal expression of $k$ contains at least $2021$ '0' digits, but the last digit of $k$ is not equal to $0$.

Find all positive integers that are zero takers.
To find all positive integers \( m \) that are zero takers, we need to ensure that there exists a positive integer \( k \) such that:
1. \( k \) is a perfect square.
2. \( m \) divides \( k \).
3. The decimal expression of \( k \) contains at least 2021 '0' digits.
4. The last digit of \( k \) is not equal to 0.

Let's break down the solution step-by-step:

1. **Constructing \( k \) with the required properties:**
   Consider \( k = (10^N + 1)^2 \) for a sufficiently large \( N \). This construction ensures that \( k \) is a perfect square because it is the square of \( 10^N + 1 \).

2. **Expanding \( k \):**
   \[
   k = (10^N + 1)^2 = 10^{2N} + 2 \cdot 10^N + 1
   \]
   This expansion shows that \( k \) will have a large number of zeros in its decimal representation. Specifically, for sufficiently large \( N \), \( k \) will have at least 2021 zeros.

3. **Ensuring \( m \) divides \( k \):**
   We need \( m \) to divide \( k \). Since \( k = 10^{2N} + 2 \cdot 10^N + 1 \), we can choose \( m \) such that \( m \) divides \( 10^N + 1 \). This ensures that \( m \) divides \( k \).

4. **Ensuring the last digit of \( k \) is not 0:**
   The last digit of \( k \) is determined by the last digit of \( (10^N + 1)^2 \). Since \( 10^N + 1 \) ends in 1, \( (10^N + 1)^2 \) ends in 1. Therefore, the last digit of \( k \) is 1, which is not 0.

5. **Generalizing \( m \):**
   From the above construction, \( m \) can be any positive integer that does not have 10 as a factor. This is because if \( 10 \mid m \), then \( m \) would force \( k \) to end in 0, which contradicts the requirement that the last digit of \( k \) is not 0.

Conclusion:
The positive integers \( m \) that are zero takers are all positive integers that are not divisible by 10.

The final answer is \( \boxed{ m } \) such that \( 10 \nmid m \).