This is a continuation of this question: A class of quadratic equations

Let $f(x,y) = ax^2 + bxy + cy^2$ be an irreducible and indefinite binary quadratic form. Consider the equation

$$\displaystyle f(x,y) = a,$$

where $a$ is the $x^2$ coefficient of $f$. Plainly, this equation is soluble in the integers since $(x,y) = (1,0)$ is a solution. Since $f$ is indefinite and irreducible, an appropriate action by the unit group of the quadratic field generated by $f$ will give us infinitely many solutions.

I am interested in the case when one can find a solution with $y$ co-prime to $a$. This is not always possible. For example, the equation

$$\displaystyle 5x^2 - 7y^2 = 5$$

has fundamental solution $(6,5)$ and all solutions are of the form

$$\displaystyle \begin{pmatrix} 6 & 7 \\ 5 & 6 \end{pmatrix}^n \binom{\pm 6}{\pm 5}.$$

The issue appears to be that $a$ divides $b$, the $xy$ coefficient of $f$. However, this divisibility property is not preserved under substitution action by $\operatorname{GL}_2(\mathbb{Z})$, so this cannot be an intrinsic obstruction.

Here is the ultimate question I am interested in. Let $O_f(\mathbb{R})$ be the maximal subgroup of $\operatorname{GL}_2(\mathbb{R})$ which fixes $f$ by substitution. It is conjugate to the split orthogonal group $O(1,1)$. Let $O_f^-$ denote the subset of $O_f$ consisting of elements of negative determinant. This part is conjugate to

$$\displaystyle \left \{ \pm \begin{pmatrix} \cosh t & \sinh t \\ -\sinh t & -\cosh t \end{pmatrix}, t \in \mathbb{R} \right \}.$$

My question is the following: for $f$ an irreducible and indefinite binary quadratic form with integer coefficients, does $O_f^-$ always contain an element in $\operatorname{GL}_2(\mathbb{Z})$? In other words, is $O_f^- \cap \operatorname{GL}_2(\mathbb{Z})$ always non-empty?

One can show that the $O_f^-$ contains elements of the shape

$$\displaystyle \begin{pmatrix} m & \dfrac{bm + cn}{a} \\ n & - m \end{pmatrix},$$

where $(m,n)$ satisfies $f(m,n) = a$. If $n$ is co-prime to $a$, then since $am^2 + n(bm+cn) = a$, it follows that $n | bm + cn$ and the above is in $\operatorname{GL}_2(\mathbb{Z})$. If $a | n$ and $a | b$, then again it is in $\operatorname{GL}_2(\mathbb{Z})$.

I've done quite a few experiments and it seems that $O_f^- \cap \operatorname{GL}_2(\mathbb{Z})$ is always non-empty.

Any help would be appreciated!