Given that, $2^{x} + 2^{-x} = 2 – (x – 2)^{2}$
$\Rightarrow 2^{x} + \frac{1}{2^{x}} = 2 – (x – 2)^{2} \quad \longrightarrow (1)$
We know that, $\text{AM} \geq \text{GM}$
$\Rightarrow \frac{2^{x} + \frac{1}{2^{x}}}{2} = \sqrt{ 2^{x} \cdot \frac{1}{2^{x}}}$
$\Rightarrow \boxed{ 2^{x} + \frac{1}{2^{x}} \geq 2}$
The minimum value of $2^{x} + \frac{1}{2^{x}}$ is $2,$ when $x=0.$
So, it follows $\text{LHS} \geq 2.$
And, $2 – (x-2)^{2} \leq 2$
$\Rightarrow – (x-2)^{2} \leq 0$
$\Rightarrow (x-2)^{2} \geq 0$
$\Rightarrow \boxed{x \geq 2}$
The maximum value of $2 – (x-2)^{2}$ is $2,$ when $x=2.$
Hence, there is no value of $x.$
Correct Answer$: \text{C}$