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.

If we let $\quad  A^2+B^2=C^2
\qquad A=3x\quad B=\sqrt{p}\quad C=y\qquad$
we can use Euclid's formula for Pythagorean triples to find infinite solutions based on the value of $(x)$.  We start with the formula
$$ \qquad A=m^2-k^2\qquad B=2mk \qquad C=m^2+k^2\qquad$$
and solve the $A$-function for $(k).\quad$
Then we test a defined range of $m$-values to see which yield integers.
\begin{equation}
A=m^2-k^2\implies k=\sqrt{m^2-A}\\
\text{for}\qquad  \lfloor\sqrt{A+1}\rfloor \le m \le \frac{A+1}{2}
\end{equation}
The lower limit ensures $k\in\mathbb{N}$ and the upper limit ensures $m> k.$
There are Pythagorean triples for all $A=2n+1$ but A=$3x$,  so we are limited to odd muliples of $3:$
$$ x\in\big\{1,3,5,7,\cdots\big\}\implies
A\in\big\{3,9,15,21,\cdots\big\}.\quad $$
Here, we will use $x=5\implies A=15$
$$A=15\implies \lfloor\sqrt{15+1}\rfloor=4\le m \le \frac{15+1}{2} =8\\
\land\quad m\in\{4,8\}\implies k \in\{1,7\} $$
$$F(4,1)=(15,8,17)\implies (x,p,y)=(5,8^2,17)\\
 F(8,7)=(15,112,113)\implies (x,p,y)=(5,112^2,113) $$
