31. $\frac{4^{x}-2^{x+1}+8}{2^{1-x}}<8^{x}$.
31. In all members of the inequality, we transition to base 2. By moving all terms to the right side of the inequality, we get $2^{x} \cdot\left(2^{2 x}+2 \cdot 2^{x}-2^{3}\right)>0$.

Since $2^{x}>0$ for any $x$, the inequality holds under the condition that $2^{2 x}+2 \cdot 2^{x}-2^{3}>0$, i.e., $2^{x}>2, x>1$.