The exponents in the prime factorization multiply to $2019$. Only $673$ and $3$ are small enough. $128^{97}=2^{679}$, so one factor must be $2^{672}$, meaning the other factor is a square less than $2^7$. Solutions are $1^2,2^2, \cdots, 11^{2}*2^{672}$

YerMom wrote: The exponents in the prime factorization multiply to $2019$. Only $673$ and $3$ are small enough. $128^{97}=2^{679}$, so one factor must be $2^{672}$, meaning the other factor is a square less than $2^7$. Solutions are $1^2,2^2, \cdots, 11^{2}*2^{672}$ So close... However, the second factor has to be the square of a prime that is not $2$, yielding only the solutions $2^{672}3^2,2^{672}5^2,2^{672}7^2,2^{672}11^2$.

Let $p$ and $q$ be distinct primes. Case 1: $n=p^{672}q^2$. We note that $128^{97}=2^{97\cdot7=679}$. We can easily see that $p=2$. So we need $q^2<2^7=128$. Thus, $q=\{3,5,7,11\}$. Case 2: $n=p^{2018}$. There are no solutions as $128^{97}<2^{2018}$. So the possible values of $n$ are \[\boxed{2^{672}\cdot3^2, 2^{672}\cdot5^2, 2^{672}\cdot7^2, 2^{672}\cdot11^2}\]