I show here that it is possible to find a same simpler result, but with no condition for the form of p.
1) If p and q=2p+1 are prime then q divides 3p+1 :
Indeed 32p-1 = (3p-1)*(3p+1) and 32p-1 = 0 [mod q] (Fermat th. for q).
If p>=5, p=6*a-1 or p=6*a+1. If p=6*a+1, q=12*a+3=0 [mod 3] , q not prime, therefore p=6*a-1 and q=12*a-1.
As q and 3 [mod 4] and 3=3 [mod 4], (3/q) = -(q/3)=-(-1/3)=1, therefore q divides 3p-1.
2) If p is prime and q=2p+1 divides 3p-1 then q is prime :
We use the "Well Known theorem" where n=h*pk+1 with h=2, p prime, n=q and k=1, it comes :
If q=2*p+1 and p>2, if there is an integer a such aq-1 = 1 [mod q] and gcd(a2-1,q) =1 then n is prime.
If a=3, 3p=1 [mod q] or 32p=3q-1 = 1 [mod q] and gcd(32-1=8,q) =1 , therefore q=2p+1 is prime.
Complements and consequences (soon ...)