Solve in $\mathbf{N}$ the equation $9x^2+p=y^2$ In these days, I have been trying to solve this problem:

Let $p \in \mathbf{N}$ a positive large integer ($> 10^9$). Find all $x, y \in\mathbf{N}$ such that:
$$9x^2+p=y^2$$

The first approach that I have tried is the following. We know that:
$$(t+n)^2-t^2=t^2+2\cdot t \cdot n +n^2-t^2=2\cdot t \cdot n +n^2$$
Also, we can build every possible square $w_n$ greater than $p$ in this way:
$$w_n=\left\lfloor\sqrt{p}\right\rfloor^2+2\cdot \left\lfloor\sqrt{p}\right\rfloor \cdot n +n^2 = \left(n+\left\lfloor\sqrt{p}\right\rfloor^2\right)^2$$
For example, let $p=5$. Follows that: $w_1=\left(1+\left\lfloor\sqrt{5}\right\rfloor^2\right)^2=9$, $w_2=\left(2+\left\lfloor\sqrt{5}\right\rfloor^2\right)^2=16$ and so on.
Now, using this idea and applying to the first equation:
$$9x^2=\left(n+\left\lfloor\sqrt{p}\right\rfloor^2\right)^2-p$$
I am not allowed to apply Pell's equation because $9=3^2$ and calcultaing $\Delta$ in $x$ doesn't help anymore.
Another approach is based on Pell's equation. I thought to express $9x^2=8x^2+x^2$ and then:
$$8x^2+x^2+p=y^2\leftrightarrow 8x^2+p=y^2-x^2\leftrightarrow (y^2-x^2)-8x^2=p \leftrightarrow u^2-8x^2=p$$
But then, in order to generate all the solutions, I have to guess the first one (or one of them) that is pretty complicated for big $p$.
So, how can we do that? Are there any other solutions?
Thanks.

An experimental approach:
Here I find one solution for this equation which a family of solutions can be based on:
$9x^2=y^2-p=(y-\sqrt p)(y+\sqrt p)$
we can can construct following system of equations:
$\begin{cases}y-\sqrt p=9\\y+\sqrt p=x^2\end{cases}$
Subtracting first equation from second one we get:
$2\sqrt p=x^2-9=(x-3)(x+3)$
Suppose:
$x-3=2\Rightarrow x=5$
$x+3=\sqrt p$
subtracting first fro, second we get:
$\sqrt p-2=6\Rightarrow \sqrt p=8\rightarrow  p=64$
Now we have:
$9x^2=y^2-64=(y-8)(y+8)$
Let $y-8=9\rightarrow y=17$
$x^2=y+8=25$
hence :
$x=5$, $y=17$ and $p=64$
$9\times5^2+64=17^2\space\space\space\space\space (1)$
$10^{12}\div 64=15625\times10^6>10^9$
So we can construct following relation by multiplying both sides of the relation (1) by $15625\times 10^6=(125\times 10^3)^2$:
$9(5\times 125000)^2+15625\times 10^6=(17\times125000)^2$
that is one solution of equation under condition $p=15625000000>10^9$ is:
$x=625\times10^3$
$y=2125\times 10^3$
This experiment shows that only with particular values of p the equation may have integer solutions.
