Find short and simple methods to solve $24x^4+1=y^2$ 
Find all  this diophantine equation  $$24x^4+1=y^2\tag{1} $$  postive integers solution

it is clear $(x,y)=(1,5)$ 
I know $y^{2}=Dx^{4}+1$, where $D>0$ and is not a perfect square, has at most two solutions in positive integers (cf. L. J. Mordell, Diophantine equations, p. 270.
Does this equation have another proof such Lucas's assertion, with short and simple methods? Like this paper: Anglin, W. S. "The Square Pyramid Puzzle." Amer. Math. Monthly 97, 120-124, 1990. The square pyramid puzzle
In the paper,Following two question have simple methods to solve it.
There are no positive integers $x$ such $2x^4+1$ is a square.
and 
There is exactly one positive integer $x$,namely $1$, such that $8x^4+1$ is a square?
But How can I find simple methods to solve $(1)$?
 A: This is simple but not elementary method.
$y^2 = 24x^4+1\tag{1}$
Using online Magma calculator as follows.
IntegralQuarticPoints($[24,0,0,0,1]$);
It says that all integral points are $[[ 0, 1 ], [ 1, 5 ], [ -1, 5 ]]$.
Hence all positive integral point is $(x,y)=(1,5).$
A: $\frac{y-1}{2} \cdot \frac{y+1}{2} = 6x^4$
So $\frac{y+1}{2} = p a^4, \frac{y-1}{2} = q b^4$ where $pq = 6$ and we need to solve the equation $p a^4 - q b^4 = 1
Case 1: $p = 6, q = 1$
This is impossible modulo 3.
Case 2: $p = 2, q = 3$
This is impossible modulo 3.
Case 3: $p = 1, q = 6$
We will show that there is no solution using the method of infinite descent. Take the minimal solution in positive integers $a, b$.
Moving sides and factoring we get $\frac{a - 1}{2} \cdot \frac{a+1}{2} \cdot \frac{a^2 + 1}{2} = 12 (\frac{b}{2})^4$, therefore there exist coprime positive integers $\alpha, \beta, \gamma$ and coprime integers $m,n,k$ such that $\frac{a-1}{2} = \alpha m^4, \frac{a+1}{2} = \beta n^4, \frac{a^2 + 1}{2} = \gamma k^4$ such that $\alpha \beta \gamma = 12$.
Notice that $\gamma | \frac{a^2 + 1}{2}$, which means that neither $2$ nor $3$ can divide $\gamma$, so we must have $\gamma = 1$ and so $\alpha \beta = 12$. Therefore, $\frac{a^2 + 1}{2} - 2 \cdot \frac{a+1}{2} \cdot \frac{a-1}{2} = k^4 - 24(mn)^4 = 1$
Now we can repeat the same argument again: $\frac{k-1}{2} \cdot \frac{k+1}{2} \cdot \frac{k^2 + 1}{2} = 3(mn)^4$, so we get $\frac{k^2 + 1}{2} = u^4$, and $\frac{k - 1}{2}, \frac{k + 1}{2}$ are equal to $3v^4, w^4$ in some order. From this we get $u^4 - 6(vw)^4 = 1$, which is a smaller solution to our original equation, a contradiction.
Case 4: $p = 3, q = 2$
In this case we have to solve the equation $3a^4 - 2b^4 = 1$. Unfortunately, I do not know of a proof of this fact which is as simple as Case 3 and entirely elementary (nor am I sure such a proof exists). However there is this paper by R.T. Bumby, which solves the more general equation $3x^4 - 2y^2 = 1$ (which has besides the trivial solution $(1,1)$ also the more surprising $(3, 11)$) using essentially elementary methods, relying only on unique factorization in $\mathbb{Z}[\sqrt-2]$.
I have only skimmed this paper by Paolo Ribenboim, but it claims to give an algorithm to find all solutions to the equation $x^2 - Dy^4 = 1$ with fixed $D$ apparently with an elementary proof.
This article is a great survey by P.G. Walsh on questions like these about Pell equations: it contains good explanations about many results and probably contains all the references you'll run across studying this topic.
Hope this answer is of help.
