Say we have two distinct integers $a$ and $b$ such that $\operatorname{gcd}(a,b)=1$ and $a < b$.
Are there an infinite number of primes of the form $a+bm$, where $m$ is a positive integer? Is this only true for some numbers $a$ and $b$?
The question title says "a set distance away from a multiple" of $b$ because the numbers in this sequence are exactly a distance b-a below their corresponding multiple of b, b(m).
Edit: All primes must be of the form a+b(m), as if a+b is prime, then gcd(a,b)=1. But are there an infinite number of primes for only certain values of a and b?
There is a theorem by Dirichlet which states exactly what you say. So there are infinitely many prime numbers with last digit 1 (or last digit 3, or 7 or 9). In fact he proves much more. Among all $\phi(b)$ arithmetic progressions with common difference $b$, the primes are equally distributed.
He proved this in 1837 bringing analysis for the first time into Number theory (quoting Tom Apostol). This is called Dirichlet's Theorem on Arithmetic Progressions. Instead of calling it an arithmetic progression let us rephrase it as below: the polynomial functions $a+bx$, with integer coefficients $a,b$ that are coprime, evaluates to primes for infinitely many integer values of $x$.
This polynomial is linear. Next ask the question about quadratic. For example, does the polynomial $x^2+1$ take prime values infinitely often. This seems to be unknown (Artin's conjecture).