Given that, $x + 1 = x^{2} \quad \longrightarrow (1)$
$\Rightarrow x^{2}-x-1 = 0 \quad \longrightarrow (2)$
Consider the quadratic equation $ax^{2} + bx + c = 0$
And the solution of the above equation is $x = \dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}$
From the equation $(2),$ we get $a = 1, b = -1,c = 1$
$x = \dfrac{-(-1) \pm \sqrt{(-1)^{2}-4(1)(-1)}}{2\times1}$
$\Rightarrow x = \dfrac{1\pm \sqrt{1+4}}{2}$
$\Rightarrow x = \dfrac{1\pm \sqrt{5}}{2}$
On neglecting the negative value (because $x>0)$, we have
$\Rightarrow \boxed{x = \dfrac{1 + \sqrt{5}}{2}}$
On squaring the equation $(1),$ we get
$({x^{2}})^{2} = (x+1)^{2}$
$\Rightarrow x^{4} = x^{2}+1+2x \quad [\because (a+b)^{2} = a^{2}+2ab+b^{2}]$
$\Rightarrow x^{4} = x + 1 + 1 +2x \quad [\because \text{From equation}\; (1)]$
$\Rightarrow x^{4} = 3x+2$
$\Rightarrow 2x^{4} = 2(3x+2)$
$\Rightarrow 2x^{4} = 6x+4 $
$\Rightarrow 2x^{4} = 6\left[\dfrac{1 + \sqrt{5}}{2}\right]+4 \quad [\because \text{Use the value of}\; x]$
$\Rightarrow 2x^{4} = 3(1+\sqrt{5}) + 4$
$\Rightarrow 2x^{4} = 3+3\sqrt{5} + 4$
$\therefore \boxed{2x^{4} = 7+3\sqrt{5}}$
Correct Answer $ : \text{D}$