Counting of natural numbers that have certain properties

How many natural numbers $$n$$ have such a property that out of all the positive divisors of number $$n$$, which are different from both $$1$$ and $$n$$, the greatest one is $$15$$ times greater than the smallest one?

After trying out different combinations, one will see, that the only ones that seem to work, are for $$n=60$$ and $$n=135$$.

Question: How to prove that there can be no others?

Hint: If $$a$$ and $$b$$ are, respectively, the smallest and the greatest divisors of $$n$$ (without $$1$$ and $$n$$ itself), then

$$a·b=n\iff a·15a=15a^2=n\implies 3\mid n$$

Can you end it now?

Hint

Note that $$n$$ must be a multiple of $$15.$$ Thus, in particular, it is a multiple of $$3.$$ So, the smallest divisor (different from $$1$$) of $$n$$ is $$2$$ or $$3.$$ Thus, the biggest divisor (different from $$n$$) of $$n$$ is $$30$$ or $$45.$$

Let's write down a factorization of $$n$$ $$n = p_1^{k_1}p_2^{k_2}...p_m^{k_m}$$ where $$p_1...p_m$$ are in increasing order.

Clearly the the smallest divisor of $$n$$ is $$p_1$$, and the greatest is $$p_1^{k_1 - 1}p_2^{k_2}...p_m^{k_m} = D$$.

So we have $$15p_1 = p_1^{k_1 - 1}p_2^{k_2}...p_m^{k_m}$$ $$3*5*p_1 = p_1^{k_1 - 1}p_2^{k_2}...p_m^{k_m}$$ Both sides are factorizations of $$D$$, therefore (by fundamental theorem of arithmetic) they are the same.

Clearly $$p_1$$ can not be greater than $$3$$ (otherwise it is not the smallest prime in factorzation of $$D$$), thus $$p \in \{2, 3\}$$

If $$p = 2$$ we have $$D = 2^{1} * 3^{1}* 5^{1}$$ and $$n = 2^{2} * 3^{1}* 5^{1} = 60$$

If $$p = 3$$ we have $$D = 3^{2}* 5^{1}$$ and $$n = 3^{3}* 5^{1} = 135$$