x=?

$\large 2^{2^{3^{2^2}}} = 2^{2^{3^4}} = 2^{2^{81}} = 2^{2\cdot 4^{40}} = 4^{4^{40}} \implies x = \boxed{40}$

Manipulating the $RHS$

$\large{4^{4^x}=(2^2)^{2^{2x}}=2^{2\cdot 2^{2x}}=2^{2^{2x+1}}}$

Substituting back

$\large{2^{2^{3^{2^2}}}=2^{2^{2x+1}}}$

$\begin{aligned} \Longrightarrow \large{3^{2^2}=2x+1} &\iff \large{3^4=2x+1}\\ &\iff \large{81=2x+1}\\ &\iff \large{x=\boxed{40}} \end{aligned}$