Solving Pell's Equation for $x^2 -7y^2 = 1$ for the first three integral solutions. Like the title states my goal is the find the first three integral solutions of the Diophantine equation. I know $x^2 -7y^2 = 1$ is a Pell's equation where $d = 7$. I found the minimal solution to be $(x,y) = (8,3)$ through brute force. I then found the next pair by using the equation:
$(x + \sqrt{7}y)^i(x-\sqrt{7}y)^i = 1$
$i = 1$ would give me the minimal solution $(8,3)$, so I went to $i = 2$.
$(x + \sqrt{7}y)^2(x-\sqrt{7}y)^2 = 1$
$\left(x^2 + 7y^2 + 2xy\sqrt{7}\right) \left(x^2 + 7y^2 - 2xy\sqrt{7}\right) = 1$ 
$(x^2 + 7y^2)^2 - 7(2xy)^2 = 1$ (This is of the form $X^2 - 7Y^2 = 1$)
I then let $X = x^2 + 7y^2$ and $Y = 2xy$, so the equation becomes the desired
$X^2 - 7Y^2 = 1$
I then plugged in my minimal solution $(8,3)$ and found $X = (8^3+7(3^3)) = 127$ and $Y = 2(8)(3) = 48$, so my new pair is $(127,48)$. To find the next solution, I let $i = 3$.
$(x + \sqrt{7}y)^3(x-\sqrt{7}y)^3 = 1$
And this is where I got stuck. I tried a few methods to get this new equation into the form of $X^2 - 7Y^2 = 1$, but have been unsuccessful. I tried expanding out the equations completely getting: 
$x^6 -21x^4y^2+147x^2y^4-343y^6 = 1$
But I'm fairly sure that's not the right direction. The other method I tried was factoring out a $(x + \sqrt{7}y)^2(x-\sqrt{7}y)^2$, so I got:
$[(x + \sqrt{7}y)(x-\sqrt{7}y)](x + \sqrt{7}y)^2(x-\sqrt{7}y)^2$ 
which simplifying I got
$(x^2-7y^2) \left( (x^2 + 7y^2)^2 - 7(2xy)^2 \right)$
I'm really not sure how to properly factor this cubic to get to the desired function form. Any help would be greatly appreciated!

Beginning with pairs $(1,0)$ and then $(8,3),$ all the other soltuions with positive variables satisfy
$$ x_{n+2} = 16 x_{n+1} - x_n \; , \;  $$
$$ y_{n+2} = 16 y_{n+1} - y_n \; . \;  $$
The $x_n$ begin
$$ 1, 8, 127, 2024, 32257, 514088, 8193151,   $$
The $y_n$ begin
$$  0, 3, 48, 765, 12192, 194307, 3096720,  $$
