I'd like to solve the following Pell equation:
$$
x^2-7y^2=-3
$$
Where $x$ and $y$ are integers. I applied the usual procedure, which avoids continued fractions:

The two minimal positive integer solutions are $(x\_0,y\_0)=(2,1)$ and $(x\_1,y\_1)=(5,2)$, thus the minimal rational solution of $x^2-7y^2=1$ should be $(p,q)=\left(\frac{4}{3},\frac{1}{3}\right)$. My script (it is in german so I don't link it here) tells me, that in this case, every pair of solutions is given by:
$$
x\_{n+1}=\frac{4}{3}x\_{n}+7\cdot\frac{1}{3}y\_n \\
y\_{n+1}=\frac{1}{3}x\_{n}+\frac{4}{3}y\_n
$$
If we proceed further, we can find that this gives:
$$
x\_n=\frac{a\_n}{3^n} \space\text{where}\space a\_0=2,\space a\_1=15,\space a\_{n+1}=8a\_n-9a\_{n-1} \\
y\_n=\frac{b\_n}{3^n} \space\text{where}\space b\_0=1,\space b\_1=6,\space b\_{n+1}=8b\_n-9b\_{n-1}
$$
But if we take these equations modulo $9$, we see that $(2,1)$ and $(5,2)$ are the only integer solution, but there surely is also $(37,14)$. Where did I go wrong? Every answer will be appreciated, but I'm not used to the approach with continued fractions, so preferably I would like to see an answer avoiding this.

EDIT:

My main question is:

Where is my fault? Or is my script wrong?