AMC 10 B Fall, 2021

The number of positive integer divisors of the positive integer whose prime factorization is $p_1^{e_1} p_2^{e_2} \cdots p_n^{e_n}$ equals $(e_1 + 1)(e_2 + 1) \cdots (e_n + 1)$. Because
$$2021 = 2025 - 4 = 45^2 - 2^2 = (45 + 2)(45 - 2) = 47 \cdot 43,$$
a number having $2021$ divisors must be of the form $p^{2020}$ or $p^{46} \cdot q^{42}$, where $p$ and $q$ are distinct primes. This is minimized by taking $p = 2$ in the first case, and $p = 2$ and $q = 3$ in the second case. Because
$$2^{46} \cdot 3^{42} < 2^{46} \cdot 4^{42} = 2^{46+2 \cdot 42} = 2^{130} < 2^{2020},$$
the least such positive integer is
$$2^{46} \cdot 3^{42} = 2^4 \cdot 2^{42} \cdot 3^{42} = 2^4 \cdot 6^{42} = 16 \cdot 6^{42}.$$
Therefore $m + k = 16 + 42 = 58$.