$x^{2} + x - 1 = 0$
Divide both side by $'x'.$
$\implies \dfrac{x^{2} + x - 1}{x} = \dfrac{0}{x}\:\:, x\neq 0$
$\implies x + 1 - \dfrac{1}{x} = 0$
$\implies x-\dfrac{1}{x} = -1$
Take square both side
$\implies \left(x-\dfrac{1}{x}\right)^{2} = (-1)^{2}$
$\implies x^{2} + \dfrac{1}{x^{2}} - 2 = 1$
$\implies x^{2} + \dfrac{1}{x^{2}} = 3$
Again take square both side
$\implies \left(x^{2}+\dfrac{1}{x^{2}}\right)^{2} = 3^{2}$
$\implies x^{4}+\dfrac{1}{x^{4}}+ 2 = 3^{2}$
$\implies x^{4}+\dfrac{1}{x^{4}} = 7$
So, the correct answer is $(C).$