I am interested in primes and some of their varying characteristics which make them 
“qualitatively” different such as whether each prime generated by a certain type of Dirichlet arithmetic progressions belongs or not to the class of primes that are obtained from powerful order terms of that progression. Particularly, I wonder whether one can go beyond Dirichlet Theorem and establish that there is an infinite supply of this sort of primes in these progressions. 


The case I am particularly interested is the case when d=1 and a=2k with k positive integer, but I would like to formulate the following **Conjecture** for the general case 


For any two coprime positive integers a and d, there are infinite primes of the form $an^2m^3+d$ with n and m integers >=0


If true this would strengthen Dirichlet theorem on arithmetic progressions selecting only the terms of the progressions whose order is a powerful number (a number that if prime p divides it, $p^2$ also divides it).


I wonder about the likeliness of the conjecture (which intuitively seems high) and whether there is any proof I am still not aware of. I welcome estimates on the toughness of the proof, as well as any hints towards its development. 



Are there infinite primes among powerful order terms of Dirichlet arithmetic progressions?
I'm sorry this isn't much of an answer, but you might be interested in this [paper](http://www.math.ualberta.ca/~subbarao/documents/Subbarao8.pdf) of De Koninck, Kátai and Subbarao. Your conjecture is almost strong enough to mean that for any squarefree $a$, there are infinitely many primes $p$ where the squarefree part of $p-d$ is exactly $a$ (the difference being that you do not require $(a,mn)=1$).


Section 4 in the cited paper establishes a sort of converse formulation: for any powerful $K$, there are infinitely many primes $p$ where the powerful part of $p-1$ is exactly $K$ (with the expected asymptotics). Of course this is a far denser set in which to look for primes, so this is certainly a much easier problem. Presumably no significant complications arise if we shift this by $d$ rather than $1$.


