Given that, ${N}^{N} = 2^{160} $

$ \Rightarrow N^{N} = \left( 2^{10} \right)^{16} $

$ \Rightarrow N^{N} = \left( 2^{5} \right)^{32} $

$ \Rightarrow N^{N} = (32)^{32} $

$\therefore \; \boxed{N = 32} $

Now, $ N^{2} + 2^{N} = 32^{2} + 2^{32} $

$ = \left( 2^{5} \right)^{2} + 2^{32} $

$ = 2^{10} + 2^{32} $

$ = 2^{10} (1+2^{22}) $

Here, $N^{2} + 2^{N}$ is a integral multiple of $2^{x}.$

So, $ 2^{x} = 2^{10} $

$ \Rightarrow \boxed{ x = 10} $

$\therefore$ The largest possible value of $x$ is $10.$

Correct Answer $:10$