Show that $13$ is the only prime such that $d(p(2p+1))=8$ and $2p+1$ is NOT semiprime where $d$ is the divisor counting function.

Question

I would like to prove the following statement.

Let $d$ be the divisor counting function and $p$ a prime number. Let $X_8$ be the set of prime numbers $p$ such that $d(p(2p+1))=8$. Then $X_8 = \{13\}\cup Y$ where $Y$ is the set of all primes $p$ such that $2p+1$ is semiprime.

Surely $13*(2*13+1) =13*3^3=351$. What I am trying to say is that $13$ is the only prime such that $d(p(2p+1))=8$ and $2p+1$ is NOT semiprime. A counter example would suffice.

I would appreciate edits to make this question more succinct and clear.

Answer 1

For a number to have 8 divisors it has to be like $p*q^3$ or $p*q*r$ where p,q,r are primes. Since you excluded semiprimes for 2p+1 then 2p+1 has to be $q^3$ meaning $$2p=q^3-1 => 2p=(q-1)(q^2+q+1)$$ and it's possible only for q=3 and p=13