The answer is no. Assume the contrary that such $f,g$ exist. Let $r_1,r_2$ be roots of $f$ and $i_1,\dots,i_6$ be such that $f(g(2^{i_k}))=0$ for $1\le k\le 6$. Let $r_1,r_2$be roots of $f$, and suppose $g(2^{i_1})=g(2^{i_2})=g(2^{i_3})=r_1$ for $i_1<i_2<i_3$. Then, $g(x)=c(x-2^{i_1})(x-2^{i_2})(x-2^{i_3})+r_1$ for some nonzero $c\in \mathbb{R}$. We then obtain that $g(2^{i_j})=r_2$ for $j\in\{4,5,6\}$. Let $T$ be the common value such that \[ h(x)\triangleq (x-2^{i_1})(x-2^{i_2})(x-2^{i_3}) = T\quad x\in\{2^{i_4},2^{i_5},2^{i_6}\}. \]Suppose $i_4<i_5<i_6$. Let $i_4<i_1<i_2<i_3$. Then, $T<0$. Since $T=h(2^{i_5})<0$, we must have either $i_5<i_1,i_2,i_3$ or $i_1<i_2<i_5<i_3$. Likewise, $i_6<i_1,i_2,i_3$ or $i_1<i_2<i_6<i_3$. So, there exists an interval from the set $(-\infty,i_1),(i_1,i_2),(i_2,i_3),(i_3,\infty)$ which contains two of $i_j$'s. This is a contradiction by comparing powers of 2 dividing $T$. Above, I assumed $i_k\ge 0$. This is without loss of generality: we can simply multiply $T$ by a large power of 2, say, $2^N$, and inspect $\tilde{i}_k=N+i_k$ instead.