Numbers with exactly 2019 positive divisors are either of the form $p^{2018}$ or $p^{672} \cdot q^2$ for distinct prime numbers $p$ and $q$. The number $128^{97}$ can be written as
$$128^{97} = (2^7)^{97} = 2^{679}.$$
As $p$ is at least $2$, the number $p^{2018}$ is greater than $2^{679}$ and therefore, the case $n = p^{2018}$ is impossible. Thus we have $n = p^{672} \cdot q^2$ with $p^{672} \cdot q^2 < 2^{679}$. Hence $p = 2$ and as $q^2 < 2^7 = 128$, $q$ is one of the primes $3$, $5$, $7$ or $11$.

Answer. There are 4 solutions: $n = 2^{672} \cdot 3^2$ or $n = 2^{672} \cdot 5^2$ or $n = 2^{672} \cdot 7^2$ or $n = 2^{672} \cdot 11^2$.