A line is given by a pair of equations: \begin{equation*} a_1 x_1 +a_2 x_2+a_3 x_3 + a_4 x_4=0, \qquad b_1 x_1 + b_2 x_2 + b_3 x_3 + b_4 x_4=0. \end{equation*} Suppose this line is on $X$. If the minor $a_3 b_4-a_4 b_3$ is non-zero, then we may rewrite the equations of the line as \begin{equation*} x_3=a x_1+ b x_2, \qquad x_4=c x_1 + e x_2. \end{equation*} Substituting into the equation of the surface $X$ we see that the expression \begin{equation*} x_1^d+x_2^d-x_2^{d-2}(ax_1 +b x_2)(c x_1 + e x_2) \end{equation*} vanishes as polynomial in x_1 and x_2. This is clearly impossible by considering the coefficient of $x_1^d$. Hence the minor $a_3 b_4-a_4 b_3=0$. So we can suppose that one of the equations of the line is of the form $a x_1 + b x_2=0$. Assume that the line does not lie on either of the planes $x_1=0$ or $x_2=0$. Thus neither of $a$ or $b$ is zero and we may rewrite this equation as $x_1=c x_2$. Substituting in the equation for $X$ we see that the line lies on the conic \begin{equation*} (1+c^d) x^2 + x_3 x_4=0. \end{equation*}\begin{equation*} (1+c^d) x_2^2 - x_3 x_4=0. \end{equation*} If $1+c^d \ne 0$, then the conic is irreducible and so does not contain a line. Hence $1+c^d=0$ and so the line is on one of the planes $x_3=0$ or $x_4=0$.
