Call a pair of integers $(a,b)$ *trivial* if it satisfies some simple divisibility condition, like for some prime $p$ we have $p$ divides both $a-1$ and $b+1$, or that $p$ divides both $a$ and $b$. This implies that $a^n+b$ is always divisible by $p$, like $7^n+2$ by $3$. But there can be more complicated conditions, like $29^n+26$, as noted by Julian in a comment. In general, $(a,b)$ is *trivial* if there is a finite collection of primes such that $a^n+b$ is always divisible by at least one prime of the collection. (Note: This has been added after Max's answer.)



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> Is it true that for every $a\ge 2$ and $b$ integers, if $(a,b)$ is non-trivial, then there are infinitely many primes in the sequence $a^n+b$?
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This would be a generalization of Fermat primes, but I couldn't find anything about this variant.
Since for $b=1$ we also know that $n$ would need to be a power of two for $a^n+b$ to be a prime, probably there are counterexamples to my question, but I don't know of any.


What happens if we also suppose $b\ge 2$?
A trivial calculation would give that the chance of $a^n+b$ to be a prime is about $\frac 1n$, so there would be infinitely many as $\sum \frac 1n=\infty$, unless there are some additional conditions, like in case of $b=1$.



Primes in shifted geometric sequence
The answer is No.


We know that $78557$ is a [Sierpinski number](https://en.wikipedia.org/wiki/Sierpinski_number) with the covering set $S = \{ 3, 5, 7, 13, 19, 37, 73 \}$, i.e., every integer of the form $78557\cdot 2^n+1$ is divisible by an element of $S$. Set $a:=2$ and $b:=128100173$, which form a non-trivial pair and satisfy
$$b \equiv 78557^{-1}\pmod{3\cdot 5\cdot 7\cdot 13\cdot 19\cdot 37\cdot 73}.$$
Then the numbers $a^n+b$ are never prime as they have the same covering set $S$.


