It is well known that a polynomial $$f(n)=a_0+a_1n+a_2n^2+\cdots+a_kn^k$$ is composite for some number $n$.
What about the function $f(n)=a^n+b$ ?
Do positive integers $a$ and $b$ exists such that $a^n+b$ is prime for every natural number $n\ge 1$ ?
I searched for long chains in order to find out whether there is an obvious upper bound. For example $4^n+4503$ is prime for $n=1,\ldots,14$.
The answer to your question is no. For any $a>1,b\geq 1$ there will always exists $n$ such that $a^n+b$ is composite.
Suppose that $a+b$ is prime, (otherwise we are finished) and consider $n=a+b$ and look at $a^{a+b}+b$ modulo $a+b$. Then by Fermat's little theorem, which states that $$a^p\equiv a\pmod{p}$$ for any prime $p$, it follows that $$a^{a+b}+b\equiv a+b\equiv 0\pmod{a+b},$$ and so $a^{a+b}+b$ is divisible by $a+b$ and hence it is composite.