Claim : $c = 0$ or $1$. Proof of Claim : From $a^n+2^n|b^n+c|b^{3n}+c^3$ and $a^n+2^n|a^{3n}+2^{3n}|b^3n+c$, we have $a^n+2^n|c^3-c \forall n \in \mathbb{N}$. Since $a^n+2^n$ becomes arbitrarily large for large $n$, we have $c^3-c=0 \implies c=0$ or $1$. Case 1: $c=0$. Let $a \neq 2$. Then, $a^n+2^n|b^n \forall n \in \mathbb{N}$. By Zsigmondy's Theorem, $\forall n >3$, $a^n+2^n$ must have a primitive prime factor. But, $\text{supp}(b^n) = \text{supp}(b)$ which is obviously a finite set. Therefore, for large enough $n$, $\exists p \in \mathbb{P}$ such that $p|a^n+2^n$ but $p \nmid b$, contradiction. Case 2: $c= 1$. Suppose that $a$ is even. Then $2^n|a^n+2^n|b^n+1 \forall n \in \mathbb{N}$. Clearly, $b$ is odd. Take $n=2$. Then, $4|b^2+1$, contradicting $b^2 \equiv 1 \pmod{4}$. Therefore, $a$ is odd. Let $\mathcal{P}$ be the set of primes in the canonical factorisation of $a$ with odd exponent. Suppose that for every prime $p$ such that $\left( \frac{2a}{p} \right) = -1$, we also have $\left( \frac{b}{p} \right) = -1$. Since $2ab$ is not a perfect square, take the sets of primes $\mathcal{P} = \{ p_1,p_2 \cdots p_r \} $ and $\mathcal{Q} = \{ q_1, q_2 \cdots q_s \}$ such that $v_{p_i}(2a)$ and $v_{q_i}(b)$ are all odd. Consider $\mathcal{A}=\mathcal{P} \setminus \mathcal{Q}$ and $\mathcal{B}=\mathcal{Q} \setminus \mathcal{P}$. Clearly, both $\mathcal{A}$ and $\mathcal{B}$ cannot be empty. Now a straightforward argument with Law of Quadratic Reciprocity, Chinese Remainder Theorem and Dirichlet's Theorem of Arithmetic Progressions gives us a contradiction. Therefore, $\exists$ some prime $p$ such that $ \left( \frac{2}{p} \right) = -\left( \frac{a}{p} \right)$ and $\left( \frac{b}{p} \right) = 1$. But then, take $n = \frac{p-1}{2}$. By Euler's Criterion, $2^n+a^n \equiv 0 \pmod{p}$. However, $b^n+1 \equiv 2 \ncong 0 \pmod{p}$, contradiction.

This is a really beautiful problem, First, I will show that $c \in ${$-1;1$} Consider a prime $p_1$ such that $\left( \frac{2a}{p_1} \right) = -1$, and placing $n=\frac{p_1-1}{2}$ we see that $p_1 \mid c^2-1$, and as there exists an infinity of such primes by quadratic reciprocity, $c^2=1$. Now I will deal with the case when $c=1$, in this case placing $n=\frac{p-1}{2}$ for some odd prime $p$ we get that $\left( \frac{2a}{p} \right) = -1 \Rightarrow \left( \frac{b}{p} \right) = -1$. Now consider the following two sets of prime divisors: $A=${$p_i \mid 2 \nmid v_{p_i}(2a)$} and $B=${$q_i \mid 2 \nmid v_{q_i}(b)$} This is not mandatory at all but will make the solution much easier to write and understand. First note that $A$ is nonempty as it contains 2, and this means that $B$ is also nonempty. I will show that $A \subset B$ and $B \subset A$ consequently. Assume that there exists a prime in $A$ which is not in $B$ say $p$. If $p$ is odd consider a prime that is $1\mod 4\frac{\prod p_i \prod q_j}{p}$ and which is an $NQR \mod p$ . This is possible by the Chinise Remainder Theorem and the Dirichlet AP Theorem. Modulo that prime we get a contradiction. And if $p$ is 2, just insdead of an $NQR$ mod $p$ make the prime to be $5 \mod 8$. Now $B \subset A$ follows by almost the same argument, just pick an odd prime which is in $B$ and not in $A$ and now again by CRT and Dirichlet, pick a prime which is an $NQR \mod a$, and $1 \mod 8$ and which also is an $NQR$ modulo a missing prime and a $QR$ modulo all the other missing primes and achieve the same contradiction. Concluding, $A$=$B$ thus $2ab$ must be a perfect square, which can not happen by the condition. I don't really know how solvable the equation is without the given constraints, but I have some progress, for example when $c=-1$ it is not hard to prove that $b$ must be a perfect square. I hope that one day I will return to the problem and find a complete solution.