If one takes in general $(\star)\, \,x^2-dy^2=c$ where $d$, $c$ in $\mathbb{N}$.

Taking $d=w^2p^2+p$ with $w\in \mathbb{Q}\ge 1$ and $p\in \mathbb{Z}$ which is verified (explained later), for the matrix $$A=\begin{pmatrix}2w^2p+1&2w(w^2p^2+p)\\2w&2w^2p+1\end{pmatrix}$$ if $X_0$ is a solution to $(\star)$ then $AX_0$ is another one.

Now $w$ could be taken in a cool way basicaly say $d=a^2b^2+cb$ with $c\in\mathbb{Z}$, $|c|<|a|$ and $c$ coprime with $b$ and $a$, letting $w=\frac{a}{c}$ and $p=cb$ the matrix $A$ is in $\mathbb{Q}$ but can have a power $A^n$ with integer entries. So to say that i didn't find any reference for this idea which is surprising. This is related and known of course as a Pell equation when $w\in \mathbb{N}$.

A question is if there is a related topic discussion to this approach since Pell equations are known, and as a conjecture to give certain family of $A$ with $A^n$ of integer entries. (It appears there are many). Thanks

Edit, i'll illustrate this in an example just for clarity: $$x^2-2021y^2=d^2$$ one solution is $(d,0)$, i took $2021$ by chance as it is within what i can get, (i don't know if it should work for $2020$) since $2021=\frac{45^2}{4^2}4^2-4$. An easy argument says if the numerator of $w$ here $45$ is $5 \pmod{8}$ Then $A^3\in \mathbb{M}_2(\mathbb{Z})$ so

$$A=\begin{pmatrix}-1011.5&45472.5\\22.5&-1011.5\end{pmatrix}$$ and $$A^3=\begin{pmatrix}-4.139590049\times 10^9&1.8609747948\times 10^{11}\\9.2081880\times 10^7&-4.139590049\times 10^9\end{pmatrix}.$$

Edit. It seems such $A$ has an all integer power $A^n$ if and only if $c$ is a power of two and mainly $|c|= 1, 2, 4$,