Solution:

A positive integer $m$ is a zero-bearer if and only if $m$ is not divisible by $10$.

Necessary condition
If $m$ is a multiple of $10$, then $m$ cannot be a zero-bearer, since every multiple of $m^{2}$ will necessarily end with at least two digits $0$.

Sufficient condition
We show that, if $m$ is not a multiple of $10$, then it is a zero-bearer. To this end, consider, for every positive integer $a$, the number
$$k = (10^{a} + 1)^{2} m^{2} = 10^{2a} m^{2} + 2 \cdot 10^{a} m^{2} + m^{2} = 10^{a} \cdot m^{2} (10^{a} + 2) + m^{2}.$$
Written in this way, the first addend is a multiple of $10^{a}$, hence its decimal expression ends with $a$ digits $0$. If we denote by $M$ the number of digits of $m^{2}$, and set $a = 2021 + M$, then we are sure that the decimal expression of $k$ will end with the $M$ digits of $m^{2}$, preceded by $2021$ digits $0$. Moreover the units digit of $k$ will be different from $0$, as required.