The only numbers with exact $2$ proper divisors are the numbers of the form $p^2$, where
p is a prime.
The proper divisors are $1$ and $p$ in this case, and $p+1$ with $p$ prime
can only be a perfect square for $p=3$.
This follows from the equation $p=a^2-1=(a-1)(a+1)$. If $a>2$ , then $p$ cannot be a prime.
So, there is only $1$ case of
$2$ divisors.
For the case of $4$ divisors, we have to find all primes $p$, such that
$p^3+p^2+p+1=(p+1)(p^2+1)$ is a perfect square.
Suppose, $q$ is a divisor of $p+1$ and $p^2+1$, so we have $p\equiv -1\ (\ mod\ q\ )$ and
$p^2\equiv -1\ (\ mod\ q\ )$.
Since we also have $p^2\equiv 1\ (\ mod\ q\ )$, we
can conclude $q=2$.
The case $gcd(p+1,p^2+1)=1$ would imply, that $p+1$ is a square, which is only
possible for $p=3$, as already mentioned, but $3^2+1=10$ is not a square.
So, we can conclude that
$$p+1=2a^2\ \ \ \ \ \ \ p^2+1=2b^2$$
with $gcd(a,b)=1$
It seems that only $p=7$ solves these equations. If there is another solution, it must
contain more than $100\ 000$ digits which I checked examining the solutions of the
equation $x^2-2y^2=-1$
The number of proper divisors is even only for squares. I checked them and
found two more examples for an even number :
$$35713^2=1275418369$$
has $8$ proper divisors.
$$102851^2=10578328201$$
has $14$ proper divisors.
Furthermore, I found an example with $3$ distinct prime factors :
$195534000$ has $399=3\times 7 \times 19$ proper divisors.
Best Answer
Hint: all divisors of an odd number are odd, so you need to have an odd number of them. Can you see why $k^2$ has an odd number of factors? The sum of divisors function is multiplicative, so you have taken care of $2^{(r+m)}$, you can consider $k$ to be odd.