Something about this is surely known. For different $m \in \mathbb Z$ we obtain different sequences $n \to2^n+m$.
Are there some $m$´s for which it is expected that this sequence should give some primes but also some computations in some relatively large ranges were done and not a single prime was found?
What are the examples of some such $m$´s?
On
Erdős had a construcion for the numbers of the form $m\cdot2^n+1$ with fixed $m$; that can be modifided for $2^n+m$. His trick was to cover the integers by aritmetic progressions.
It is easy to verify that $3|2^2-1$, $7|2^3-1$, $5|2^4-1$, $17|2^8-1$, $13|2^{12}-1$ and $241|2^{24}-1$. The set of positive integers can be covered by the arithmetic progressions $\{2k+1\}$, $\{3k+1\}$, $\{4k+2\}$, $\{8k+4\}$, $\{12k+8\}$ and $\{24k\}$.
Hence, let $M=3\cdot5\cdot7\cdot13\cdot17\cdot241=5592405$, and choose a positive integer $m>241$ such that $3|m+2^1$, $7|m+2^1$, $5|m+2^2$, $17|m+2^4$, $13|m+2^8$ and $241|m+2^{0}$. Such an $m$ exists due to the Chinese Remainder Theorem. Then every number of the form $2^n+m$ is divisible by either $3,5,7,13,17$ or $241$.
$m=2131$ seems to be a hard case. $n=316$ and $n=496$ show that no small factor is forced. On the other hand, $$2^n+2131$$ is not prime for $1\le n\le 40\ 000$
$m=2\ 491$ gives a prime for $n=3\ 536$ and $4\ 471$ gives a prime for $n=33\ 548$
I can continue the search of hard cases in the case of interest.
"Survivors" upto $n=1\ 000$ in the range $[-10^5,10^5]$ (I omitted the even $m$)
$355$ survivors
$10^4$-survivors in the range $[-10^5,10^5]$
$117$ survivors