$y= (x+\sqrt{x^2-1})^m\\ \dfrac{dy}{dx}=y_1=m. (x+\sqrt{x^2-1})^{m-1}.\bigg[1+\dfrac x{\sqrt{x^2-1}}\bigg]\\ y_1=m\dfrac{(x+\sqrt{x^2-1})^m }{\sqrt{x^2-1}}\\ \therefore y_1= \dfrac{my}{\sqrt{x^2-1}} , y_1\sqrt{x^2-1}=my\\ y_2=m\Bigg[\dfrac{\sqrt {x^2-1}y_1-\dfrac x{\sqrt{x^2-1}}y}{(x^2-1)}\Bigg].\\ y_2=m\bigg[\dfrac{my-\dfrac {y_1}m.x}{x^2-1}\bigg]\\ y_2(x^2-1)=m^2y-xy_1$

$y_2(x^2-1)+xy_1 -m^2y =0\\ Applying\ the \ Leibnitz\ Theorem \\ (x^2-1).y_{n+2}+nx(2)y_{n+1}+{n(n-1)}y_n-xy_{n+1}-ny_n-m^2y_n=0\\ (x^2-1)y_{n+2}+x(2n+1)y_{n+1}+(n^2-m^2)y_n=0 $

$(x^2-1).y_{n+2}+nx(2)y_{n+1}+{n(n-1)}y_n-xy_{n+1}-ny_n-m^2y_n=0\\ (x^2-1)y_{n+2}+x(2n+1)y_{n+1}+(n^2-m^2)y_n=0--- Proved$