Numbers $m = pq^4$ ($p,q$ are distinct primes) for which $m$ divided by the number of its factors is an integer

Question

The $\operatorname{Ionof}$ (Integer on number of factors) of an integer is the integer divided by the number of factors it has. For example, $18$ has $6$ factors so $\operatorname{Ionof}(18) = \frac{18}{6} = 3$, and $27$ has $4$ factors so $\smash{\operatorname{Ionof}(27) = \frac{27}{4} = 6.75}$.

This has been asked here --- Solving a Word Problem relating to factorisation --- but it was not given context.

If $p$ and $q$ are distinct primes, find all numbers of the form $pq^4$ whose $\operatorname{Ionof}$ is an integer.

I have tried substituting $p$ and $q$ for real numbers, but have not seen any consistent pattern.

I am really at a loss here. Any hints would be helpful, and also let me know if I need to reword, or provide more context.

Answer 1

$pq^4$ has $10$ factors: $1, p, q, q^2, q^3, q^4, pq, pq^2, pq^3, pq^4$. This means that for Ionof$(pq^4)$ to be an integer, $pq^4$ must be divisible by $10$. Then $p$ and $q$ must multiple to a factor of $10$. The only prime numbers which serve this are $5$ and $2$. Therefore $p=2, q=5$, or $p=5, q=2$.

Answer 2

Hint If $n \mid pq^4$, then the prime factorization of $n$ contains only factors of $p$ and $q$ (no more than one of the former and four of the latter). How many such numbers are there?