Do surface groups embed into PSL_2 over a real quadratic integer ring? $\DeclareMathOperator\PSL{PSL}$ Let $ \mathbb{Z} $ be the ring of integers and $ \mathbb{R} $ the field of real numbers. Let $ \Sigma_g $ be a surface of genus $ g \geq 2 $. Let $ \pi_1(\Sigma_g) $ be the fundamental group of the surface. There are many way to embed $ \pi_1(\Sigma_g) $ into $\PSL_2(\mathbb{R}) $. There are, however, no ways to embed $ \pi_1(\Sigma_g) $ into $ \PSL_2(\mathbb{Z}) $. Given some $ g \geq 2 $, is there a good way (an algorithm) to find a real algebraic integer $ \alpha $  such that $ \pi_1(\Sigma_g) $ embeds in $ \PSL_2(R) $? Here $ R $ is the ring
$$
R:=\mathbb{Z}[\alpha]
$$
(Preferably $ \alpha $ is just a (real) quadratic extension. That is, $ \alpha $ is the root of some polynomial $ x^2+bx+c $ where $ b^2-4c \geq 0 $ and $ b,c \in \mathbb{Z} $.)
History of the question: The original question claimed that surface groups do not embed in $ \PSL_2(\mathbb{Q}) $ and asked for embeddings into $ \PSL_2(\mathbb{F}) $ where $ \mathbb{F} $ is a finite degree field extension of $ \mathbb{Q} $. The claim that surface groups do not embed in $ \PSL_2(\mathbb{Q}) $ is false. In fact there are many such embeddings.
The first edit of the question fixed this and asked instead for an embedding into $ \PSL_2(R) $ for $ R $ a finite rank extension of $ \mathbb{Z} $ by algebraic integers. The current version of the question is the second edit.
 A: For the new question an answer is given by arithmetic Fuchsian groups. For example it is well-known that the reflection group associated with the regular right-angled pentagon in $\mathbb H^2$ contains every surface group as it contains the fundamental group of the non-orientable surface of Euler characteristic -1 as an index-4 subgroup. On the other hand it is an index-10 subgroup in the (2,4,5)-triangle group $\Delta$ (by dividing the pentagon into triangles from the center with vertices on the pentagon's and on the middles of its edges). The latter is known to be arithmetic by a result of Takeuchi (see for example item 6 in the table in section 13.3 of Maclachlan--Reid https://zbmath.org/?q=an%3A1025.57001). Its trace field is $\mathbb Q(\sqrt 5)$, which means that $\Delta$ will be realisable as a subgroup of $\mathrm{PSL}_2$ over a quadratic extension $F$ of $\mathbb Q(\sqrt 5)$, and contained in the ring of integers $\mathbb Z_F$. There are many choices for such an extension, for example one can take $F = \mathbb Q(\sqrt 2, \sqrt 5)$ since the quaternion algebra ramifies only at primes dividing 2. So in principle you get an embedding of $\Delta$, and hence of any surface group, into $\mathrm{PSL}_2(\mathbb Z_F)$. You can ask Sage to compute an integral basis for the latter to get a complete answer to your question.
Another possibility to avoid the arithmetic machinery would be to use hyperbolic geometry to compute generators for the triangle group directly and see where they lie, this has probably been done but I don't know a reference (and I'm not sure how to choose the walls as to get something which would lie where it should, i.e. $\mathbb Z_F$).
This may be not optimal, maybe all surface groups embed into $\mathrm{PSL}_2$ over a real quadratic field. For example the (2, 4, 6)-triangle group is arithmetic with trace field $\mathbb Q$, it embeds into $\mathrm{PSL}_2(\mathbb Z[\sqrt 2])$ if I'm not mistaken (see the table in Maclachlan--Reid) so its surface subgroups do as well. However I don't know which genera will be realised this way.

EDIT For the new question a positive general answer is given by computations of John Voight which he recorded in https://arxiv.org/pdf/0802.0911.pdf. The paper gives a complete list of all "Shimura curves" (a particular family of hyperbolic 2-orbifolds) whose underlying surface has genus at most 2. In particular his table 4.1 indicates that the  curve associated to the full unit group of the unique maximal order in the $\mathbb Q$-quaternion algebra of discriminant 26 has genus 2 and no singularities, so the image of this group in $\mathrm{PSL}_2(\mathbb R)$ is isomorphic to the surface group of genus 2 and it is contained in $\mathrm{PSL}_2(\mathbb Z[\sqrt 26])$.
This representation should be computable explicitely using software developed by Voight and others, that i'm not too familiar with (i don't think it's available in Sage currently).
A: Here is, at Moishe Kohan's request, an optimal answer to the original question (which asked for a representation over any number field; as i noted in the comments there exists plenty of surface group representations with rational coefficients, as follows from a theorem of Takeuchi).

To prove the existence of single a $\mathrm{PSL}_2(\mathbb Q)$-representation of a surface group is quite simpler than the full proof of Takeuchi's theorem (which is not super hard itself in the torsion-free case). Of course it suffices to prove it in genus 2. To do so observe first that it is possible to find two matrices $A_1, B_1 \in \mathrm{SL}_2(\mathbb Q)$ such that they generate a discrete subgroup of $\mathrm{SL}_2(\mathbb R)$ and
$$ A_1B_1A_1^{-1}B_1^{-1} = \left(\begin{array}{cc} a & 0 \\ 0 & a^{-1} \end{array}\right) $$
for some $a \in \mathbb Q \setminus \{0, \pm 1\}$ and the quotient $\langle A_1, B_1 \rangle \backslash \mathbb H^2$ is a one-holed torus (the set of matrices $A, B$ satisfying the last condition is an open set, so rational matrices are dense there, and we can always conjugate to diagonalise the commutator over $\mathbb Q$). Geometrically, in the half-plane model, this means that a fundamental domain for the action of $\langle A_1, B_1 \rangle$ on the convex hull of its limit set is a polygon with one side on the geodesic $c$ the geodesic from 0 to $\infty$ and the adjacent sides are orthogonal to $c$; the boundary component of the quotient torus (the convex core of $\langle A_1, B_1 \rangle \backslash \mathbb H^2$, let us call it $T$) is the image of $c$.
Now let $\sigma$ be the reflexion in $c$, and $A_2, B_2$ be the conjugates of $A_1, B_1$ by $\sigma$. Then $\langle A_2, B_2 \rangle \subset \mathrm{PSL}_2(\mathbb Q)$, and the quotient of $\mathbb H^2$ by $\Gamma = \langle A_1, B_1, A_2, B_2 \rangle$ is the double of $T$, that is a genus 2 surface. Hence $\Gamma$ is a genus 2 surface group inside $\mathrm{PSL}_2(\mathbb Q)$ which is discrete in $\mathrm{PSL}_2(\mathbb R)$.

Takeuchi's argument for the full theorem is similar to this, except using Calabi--Weil rigidity instead of the geometric argument, and being more precise about commutators. It is also possible to prove it using Fenchel--Nielsen coordinates.
