My approach so far:
The given equation can be rewritten as: $x^2 -y^2=\frac{y^2 -1}{24}.$ This gives
$24x^2 +1=25y^2=(5y)^2.$ So $(24x^2+1)$ must also be a perfect square. This implies $x=0, 1$ is two such possible values for $x$. As $x>0$, so $x=1$ is a possible solution. Corresponding to this, we get $y=1$. But how to check does there exist any other solutions or not? Please suggest.. Thanks in advance.
On
$x^2-Dy^2=1$
Let $(x_0,y_0)$ is the smallest nontrivial positive solution.
Then general solution is given as follows.
$x_n + y_n\sqrt{D} = (x_0 + y_0\sqrt{D})^n$
Let $X=5y$ and $Y=x$ then $24x^2 - 25y^2 =- 1$ is transformed to $X^2-24Y^2=1.$
Smallest nontrivial positive solution is $(X0,Y0)=(5,1).$
Then all solutions are given $X_n + Y_n\sqrt{D} = (5 + 2\sqrt{6})^n.$
We show ten solutions with $n=1\cdots 10.$
$[5+2\sqrt{6}], [49+20\sqrt{6}], [485+198\sqrt{6}], [4801+1960\sqrt{6}], [47525+19402\sqrt{6}], [470449+192060\sqrt{6}], [4656965+1901198\sqrt{6}], [46099201+18819920\sqrt{6}], [456335045+186298002\sqrt{6}], [4517251249+1844160100\sqrt{6}]$
Since $X_n$ must be divisible by $5$, we know $[485+198sqrt{6}]$ is second solution.
Thus $y=\frac{485}{5}=97, x=\frac{198}{2}=99.$
Third solution is $y=\frac{47525}{5}=9505, x=\frac{19402}{2}=9701.$
The equation is equivalent to $$ 24x^2 - 25y^2 =- 1. $$ An equation of the form $ax^2-by^2=c$ can be solved by continued fractions, see here, or by the LMM method. The solutions are given by the family $$ x = u + 25 v, y = u + 24 v $$ with $u^2-600v^2=1$. This is Pell's equation. Its fundamental solution is $(u,v)=(49,2)$, which gives $(x,y)=(99,97)$. So we have infinitely many solutions.