Is It more, less or equal and might be nothing

m= $({ \log _{ 2 }{ 5 } ) }^{ 2 }$
Let $({ \log _{ 2 }{ 5 } ) }^{ 2 }$ =x
so by simplifying n= $\log _{ 2 }{ 20 }$
we get n= $2+\log _{ 2 }{ 5 }$
or n= 2+x
At x=2 we get m=n but if x>2 then
${ x }^{ 2 }$ >x+2
And we know that $\log _{ 2 }{ 5 }$ is greater than 2 so
$\boxed{m>n}$

Note that $n = \log_2 20 = \log_2 4 + \log_2 5 = 2+\log_2 5$ . Also that

$\begin{aligned} m & = (\log_2 5)^2 \\ & = \log_2 5 \times \log_2 5 & \small \color{#3D99F6} \text{Note that } \log_2 5 > \log_2 4 = 2 \end{aligned}$

$\implies m > 2 \log_2 5 = {\color{#3D99F6}\log_2 5} + \log_2 5 > {\color{#3D99F6}2} + \log_2 5 = n$

$\implies \boxed{m>n}$