by zsignmondy $a^{2}+4^{b}$ divisible by its primitive divisor p $n^{2023} \equiv 0 \mod p$ meaning n is divisible by p however $p \nmid a+2^b$ so no such sols

I'm not sure if the above solution is correct, but here is a simpler solution: If $n= 1$ then there doesn't exist $a,b$ If $n\ge 2$: Let's take $p$ as a prime divisor of $n$ Then $p|a+2^b$ and $p|a^2 + 4^b=(a+2^b)^2-4a \cdot 2^b$ Therefore $p|a+2^b$ and $p|4a\cdot 2^b$ Now we consider when $p=2$ and $p>2$ If $p>2$, then $p|a+2^b$ and $p|a.\cdot 2^b$, meaning that $a$ and $2^b$ are both divisible by $p$, which implies $p=2$, absurd!! From here we also imply that $n$ can only be a power of $2$. Now assume $v_2(n)=x$ and $v_2(a)=y$. We have: $v_2(a^2+4^b)=min\{v_2(a^2),v_2(4^b)\}=min\{2y,2b\}$ $v_2(n^{2023})=2023x$ Therefore $2023x = min\{2y,2b\}$ Doing similarly we also imply that $2022x=min\{y,b\}$ Now we consider when $y > b$ (the other cases can be done similarly), then $2023x=2b,2022x=b$, so x = 0, absurd!! So there are no solutions

$2n^{2023} =2*(a^2+4^b) \geq (a+2^b)^2=n^{4044} \to n^{2021} \leq 2$ so $n=1$ but for $n=1$ there are not solutions

nice problem,beautiful solutions!