At first glance, I think this problem be relevant to Zsigmondy Theorem.
By set $2^{n} + 3 =k$ and then become $2^{n} -1^{n} = k-4$
Which $2^{n} -1^{n} $ has at least one primitive prime factor except $2^{6} -1^{6} $
Am I on the right way? , to prove there are infinite $n$ which $2^{n} +3 $is composite.
Thank you and I appreciate any reply.
On
Hint $$2^{3n+2}+3 \equiv 0 \pmod{7} \\ 2^{10n+3}+3 \equiv 0 \pmod{11} \\ 2^{18n+4}+3 \equiv 0 \pmod{19} \\ 2^{66n+6}+3 \equiv 0 \pmod{67} $$
and so on. As longa s you find ONE solution $(a,p)$ such that $$2^a+3 \equiv 0 \pmod{p}$$ for some prime $p$ you get for free that $2^{(p-1)n+a}+3$ is divisible by $p$ and hence composite for all $n$.
You can always find such $(a,p)$ by picking an $a$ at random and picking $p|2^a+3$.
It's a wierd problem.
My argument would be: if $n$ is of the form $4k + 1$, then $2^n + 3$ is divisible by $5$.
Hence composite for all $k > 0$.