While prime factorization doesn't exactly happen instantly for large numbers, you can determine the number of divisors of a number using the exponents from the prime factorization.
For every distinct power in a number's prime factorization, there is an integral exponent. Add 1 to every one of these powers, multiply the resulting numbers, and you will obtain the number of divisors that divide even into the original number.
Some examples are called for to demonstrate the power of this fact (of which I can't recall a rigorous proof).
Example 1: 60. The prime factorization of 60 is $2^23^15^1$. Just looking at each of the exponents, we have 2,1,1. Add one to each of them to get 3,2,2 and multiply them to get 3*2*2 = 12. This means that 60 has 12 divisors (including 1 and 60 itself). Indeed, the positive divisors of 60 are, in ascending order, 1,2,3,4,5,6,10,12,15,20,30,60.
Example 2: 17. 17 is prime, so its prime factorization is (explicitly with exponents) 17^1. Take the exponent 1, add one to it to get 2, and now you have the number of positive divisors of 17: 1,17.
Example 3: $p$, where $p$ is any prime. Needless to say, the prime factorization is $p^1$, and the positive divisors of $p$ are 1,$p$. Since the number of divisors is always 2 (which is prime), we can conclude that all primes have a prime number of divisors.
Now we can consider composites. Every natural number $n\geq 1$ must have 1 and $n$ at least in its prime divisor list. For every other divisor $d$ of $n$, there is a corresponding divisor $d/n$ that keeps the number of divisors even. The only special case is when $d=\frac{n}{d} \implies d^2 = n$. But then $d$ is just the square root of $n$ (!) and will only occur once in the list of $n$'s divisors, thus producing an odd number of divisors. So the only composite numbers with an odd number of divisors are perfect squares. There is no need to check any non-square composite number because they will have an even number of at least 4 divisors.
Computational Pro-Tip: when searching for all numbers on the interval $[a,b]$, "simply" list the primes in the interval. Then assign $m = \lfloor \sqrt{a} \rfloor$, and $n = \lceil \sqrt{b} \rceil$. Find the prime factorization for every natural number on the interval $[m,n]$, double each of the exponents in said prime factorization, add one to every exponent, multiply all of them as before and determine if this product is prime. If so, just append the square of corresponding integer to the list that keeps track of the numbers with a prime number of positive divisors.
I can think of one extreme case, numbers much larger than you are using. Given a real number $\delta > 0,$ the exponent of some prime $p$ in the superior highly composite number associated with $\delta$ is
$$ k = \left\lfloor \frac{1}{p^\delta - 1} \right\rfloor. $$
For small primes these exponents are roughly proportional to $1 / \log p,$ so they decrease fairly quickly at first, in any case they never increase, and no prime is skipped over. These numbers, and all highly composite numbers, are the product of primorials.
For example, taking $\delta = 0.085$ gives us the S.H.C. number
$$ N_{0.085} = 2^{16} \cdot 3^{10} \cdot 5^6 \cdot \mbox{more}. $$
The number of divisors is
$$ d(N_{0.085}) = 17 \cdot 11 \cdot 7 \cdot \mbox{more}. $$
This number is divisible by $17$ but not by $13,$ hence it cannot be a highly composite number.
Taking $\delta = 0.05$ gives us the S.H.C. number
$$ N_{0.05} = 2^{28} \cdot 3^{17} \cdot 5^{11} \cdot \mbox{more}. $$
The number of divisors is
$$ d(N_{0.05}) = 29 \cdot 18 \cdot 12 \cdot \mbox{more}. $$
This number is divisible by $29$ but not by any of $23,19,17,13,$ hence it cannot be a highly composite number.
We can invert things: given a prime $p$ and a target exponent $k,$ the largest $\delta$ that will assign the exponent $k$ to the prime $p$ is
$$ \delta = \frac{\log \left( \frac{k+1}{k} \right)}{\log p} $$
So, you see, it is possible to generate S.H.C. numbers on your own, and prescribe one of the exponents as you see fit. Note that the exact value, which will cause numerical problems, is not necessary; my preference is to take $\delta$ a rational number with
$$ \frac{\log \left( \frac{k+1}{k} \right)}{\log p} > \delta > \frac{\log \left( \frac{k+2}{k+1} \right)}{\log p}, $$ which still gives the correct $k.$
I wanted the exponent of $2$ to be a prime minus $1,$ but the exponent of $3$ to be smaller than the previous prime minus $1;$ you can see how I did that now.
Edit: factoring; the ratio of consecutive S.H.C. numbers is a single, very small, prime. If you have factored one of these, you can factor the next one by taking the ratio, and just correcting the appropriate exponent by $1.$ Often, the exponent increased is for a new prime, exponent bumped from $0$ to $1.$ Anyway, you can keep doing this to factor all of them that anyone has listed, such as http://oeis.org/A002201 and text list here http://oeis.org/A002201/b002201.txt Oh, not by the way, the increased exponent found by ratio tells you what $\delta$ can be for the larger S.H.C. number.
Best Answer
If the prime factorization of $N$ is $\prod_i p_i^{\alpha_i}$ where the $p_i$s are all different, then the total number of divisors is $\prod_i (\alpha_i+1)$, and the number of proper divisors is 1 less than that.
So it is trivial to manufacture a number with any desired number of proper divisors -- nothing forces it to be prime.
The hardest number of proper divisors to reach is seen to be ones that are a prime minus one; they can only be reached by having the original number be a prime power.
It is dangerous to try to conclude anything from a table that goes op to 20 only; unusually many among these first numbers are prime, and you're not going far enough to reach a high power of anything.