Can exist an even number greater than $36$ with more even divisors than $36$, all of them being a prime$-1$?

Suppose that $n$ has the property that $d+1$ is prime for each even divisor $d$ of $n$, and let $k$ be the number of powers of $2$ that divide $n$.

• If $k=0$, there is nothing to say.

• $k\geq 3$ is impossible, since $2^3 + 1$ is not prime.

• Suppose that $k=2$, and let $p$ be an odd prime divisor of $n$. Then $2p+1$ and $4p+1$ are prime. But working modulo $3$, we can see that this is only possible if $p=3$. So $n=4\cdot 3^l$ for some $l\geq 0$. But $2\cdot 3^3 + 1$ is not prime, so the only possibilities are $n=4,12,36$.

The remaining case is $k=1$. If $p\geq 5$ is a prime factor of $n$, then $p^2$ does not divide $n$, because $2p^2+1$ is divisible by $3$. As above, $3^3$ cannot divide $n$, so $n=2\cdot 3^l\cdot \prod_{i=1}^t q_i$, where $0\leq l \leq 2$ and the $q_i$ are distinct primes $\geq 5$.

• If $t=0$, we have the cases $n=2,6,18$.

• If $t=1$, we have the cases $n=2p, 6p, 18p$, none of which have more than six even factors.

• $t\geq 2$ is impossible: at least one of $2p+1$, $2q+1$, and $2pq+1$ must be divisible by $3$.

We conclude that $n$ cannot have more than six even factors, and it has precisely six only when $n=18p$ for some prime $p$.

Sieving with small primes reveals the following.

Assume that $$ n=2^a\cdot3^bp_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}\qquad(*) $$ has the property that $d+1$ is a prime whenever $d$ is an even factor of $n$. Here $(*)$ gives the prime factorization of $n$, so $p_i$ are all distinct primes $>3$. Without loss of generality we can assume that $a>0$ and that $a_i>0$ for all $i$.

If any of the primes $p_i$ satisfies the congruence $p_i\equiv1\pmod 3$, then $2p_i+1$ is divisible by three, and $d=2p_i\mid n$ is in violation of the assumption. We can conclude that $p_i\equiv-1\pmod 3$ for all $i$.

If $k\ge 2$ then $2p_1p_2+1$ is divisible by three, so $d=2p_1p_2$ is in violation. Therefore $k\le1$. But also if $a_1\ge2$, then $2p_1^2+1$ is divisible by three and $d=2p_1^2$ is a violating factor.

At this point we know that either (call this case A) $$ n=2^a\cdot3^bp $$ for some prime $p\equiv-1\pmod3$, or (call this case B) $$ n=2^a3^b. $$

In case A we make the following further observations. First we see that $4p+1$ is again divisible by three, so $d=4p$ is in violation. Therefore in case A we must have $a=1$. Also, $2\cdot 3^3+1$ is a multiple of five, so we similarly conclude that $b\le 2$.

In case B we observe that $2^3+1$ and $2\cdot 3^3+1$ are no primes, and therefore $a\le2$ and $b\le2$.

So we are left with the possibilities

• $n=2\cdot 3^b p$ with $b\le2$. This $n$ has $2(b+1)$ even factors, so there are at most six of them.
• $n=2^a\cdot 3^b$ with $a,b\le2$. This $n$ has $a(b+1)$ even factors, so again there are at most six of them.

I made a program that calculates numbers, that you are searching, but only found $$198,26118,347\color{brown}58,49338,67698,79038,109818,...$$ which has only $6$ '$prime-1$' divisors like $36$. The range was $10^8$.

It shows that a number with even, $7$ '$prime-1$' divisors like these does not exist or they are very big to calculate...