Is there a well known proof that shows solutions of $y^2=3x^4+1$ are only (1,2), (2,7) over positive integers?

Question

I found a theorem from a book 'Diophantine equations', L. J. Mordell, which says

The equation $y^2 = Dx^4+1$ where $D>0$ and is not a perfect square, has at most two solutions in positive integers.

But I can't find any proof in this book, and I tried to find its proof but I failed. Is there anybody knows its proof?

Answer 1

$y^2 = 3x^4+1\tag{1}$
Using online Magma calculator as follows.
IntegralQuarticPoints($[3,0,0,0,1]$);

It says that all integral points are $( ( 0, 1 ), ( 1, 2 ), ( -1, 2 ), ( 2, 7 ), ( -2, 7 ))$.
Hence all positive integral points are $(x,y)=(1,2),(2,7).$

Answer 2

$$y^2 = Dx^4+1\implies x = ± \frac{\sqrt[4]{y^2 - 1}}{\sqrt[4]{D}}\implies x^4=\frac{y^2-1}{D}\quad x,y\in\mathbb{N}$$

We can see that $y^2-1$ must be a $0$ or multiple of $(D)$ and $(y^2-1)/D$ must also be a $4^{th}$ power integer. Simple spreadsheet formulas reveal $(D,x,y)$ values as shown in these samples where $(D,0,1)$ is valid for all values of $D$ but omitted here for brevity.

$$ (3,1,2)\quad (3,2,7)\quad (5,2,9)\quad (8,1,3)\quad (14,2,15)\quad (15,1,4)\quad (18,2,17)\quad (24,1,5)\quad (33,2,23)\quad (35,1,6)\quad (39,2,25)\quad (48,1,7)\quad (60,2,31)\quad (63,1,8)\quad (68,2,33)\quad (80,1,9)\quad (99,1,10)\quad ... $$