Highest Possible Number for $x$ (Part 2)

$\begin{aligned} 2^{3^4} \div 4^x & = y \\ \frac {2^{3^4}}{4^x} & = y \\ 2^{3^4} & = 4^x\color{#3D99F6}y & \small \color{#3D99F6} \text{Let }y = 2^n \\ 2^{81} & = 2^{2x}\color{#3D99F6}2^n \\ 2^{81} & = 2^{2x+n} \\ \implies 2x + n & = 81 \end{aligned}$

Note that $y$ is an integer if $n$ is an integer. Then the largest integral $x$ is $\boxed{40}$ , when $n=1$ and $y=2^1=2$ .

(2^81)/(2^(2x)) = y , or 2^(81 - 2x) = y, so x = 40 is the largest integer, y = 2.