If $2^{2^j} a + 1$ divides $c^{2^j}+1$ for fixed $a,c$ and all $j$, then $a=1$,$c=2^l$ for some odd $l$, or $a=0$.

As Elaqqad said in his comment, we can use Hadamard Quotient Theorem. For example, in this article (in french), it is proved the following result (see page 190).

Theorem : if $u$ and $v$ are in $\mathbb{Z}\backslash\{-1;0;1\}$ and $P_0$, $P_1$, $Q_0$, $Q_1$ are polynomials in $\mathbb{Z}[X]\backslash\{0\}$ such that for all integer $n>0$, $$P_1(n)u^n+P_0(n)\,|\,Q_1(n)v^n+Q_0(n)$$ then we can find $t\in\mathbb{N}^*$ such that $v=u^t$ and $Q_0P_1^t=(-1)^{t+1}Q_1P_0^t$.

So it is easy to deduce that if $a\not=0$ is such that $2^na+1\,|\,b^n+1$ for all integer $n>0$, then $a=1$ and $b=2^{2k+1}$ for some $k\geq0$.