I would guess that as soon as you have more than three cusps, the Teichmuller space associated to the surface is non trivial (a complex manifold with dimension > 0) whereas there are countably many (up to isometry) surfaces whose associated group is commensurable to $PSL_2(\mathbf{Z})$. So most discrete groups associated to a surface with sufficiently many cusps (4 probably) are not commensurable to $PSL_2(\mathbf{Z})$.

All discrete subgroups of $PSL_2(\mathbf{R})$ with finite covolume are obtained as follows. Choose a 2n-gone in the hyperbolic Poincare half-plane all of whose vertices are at infinity and such that opposite sides are identified through an isometry of the half-plane. The group generated by these isometries has the polygon as its fundamental domain, and since its vertices are at infinity, this fundamental domain has finite volume. All finite covolume discrete groups without elliptic elements are obtained this way. This is a particular case of Poincare theorem for surfaces, that shows that there is a correspondance between finitely generated subgroups of $PSL_2(\mathbf{R})$ and convex fundamental polygons in the half-plane. See Beardon, "The Geometry of Discrete Groups", chapter 9. As soon as you have more than 6 vertices, you have a free choice for a pair of points that you cannot compensate by a global isometry (that operates triply transitively on the half-plane but no more). So you have uncountably many different subgroups that are of finite covolume and no two of them are conjugated.

I would guess that as soon as you have more than three cusps, the Teichmuller space associated to the surface is non trivial (a complex manifold with dimension > 0) whereas there are countably many (up to isometry) surfaces whose associated group is commensurable to $PSL_2(\mathbf{Z})$. So most discrete groups associated to a surface with sufficiently many cusps (4 probably) are not commensurable to $PSL_2(\mathbf{Z})$.

All discrete subgroups of $PSL_2(\mathbf{R})$ with finite covolume are obtained as follows. Choose a 2n-gone in the hyperbolic Poincare half-plane all of whose vertices are at infinity and such that opposite sides are identified through an isometry of the half-plane. The group generated by these isometries has the polygon as its fundamental domain, and since its vertices are at infinity, this fundamental domain has finite volume. All finite covolume discrete groups without elliptic elements and with zero genus are obtained this way. This is a particular case of Poincare theorem for surfaces, that shows that there is a correspondence between finitely generated subgroups of $PSL_2(\mathbf{R})$ and convex fundamental polygons in the half-plane. See Beardon, "The Geometry of Discrete Groups", chapter 9. As soon as you have more than 6 vertices, you have a free choice for a pair of points that you cannot compensate by a global isometry (that operates triply transitively on the half-plane but no more). So you have uncountably many different subgroups that are of finite covolume and no two of them are conjugated.

EDIT: to be slightly more precise in the case of non-zero genus, all discrete groups with finite covolume are obtained by identifying sides of a finite-sided convex polygon that touches the boundary in finitely many points, and such that the angles at the vertices inside the disk that are glued together sum up to $2\pi$, or a submultiple of $2\pi$ if the vertices are fixed by elliptic elements. And all such side-paired polygons generate a discrete subgroup of finite covolume.

I would guess that as soon as you have more than three cusps, the Teichmuller space associated to the surface is non trivial (a complex manifold with dimension > 0) whereas there are countably many (up to isometry) surfaces whose associated group is commensurable to PSL_2(Z). So most discrete groups associated to a surface with sufficiently many cusps (4 probably) are not commensurable to PSL_2(Z).

All discrete subgroups of PSL_2(R) with finite covolume are obtained as follows. Choose a 2n-gone in the hyperbolic Poincare half-plane all of whose vertices are at infinity and such that opposite sides are identified through an isometry of the half-plane. The group generated by these isometries has the polygon as its fundamental domain, and since its vertices are at infinity, this fundamental domain has finite volume. All finite covolume discrete groups without elliptic elements are obtained this way. This is a particular case of Poincare theorem for surfaces, that shows that there is a correspondance between finitely generated subgroups of PSL_2(R) and convex fundamental polygons in the half-plane. See Beardon, "the geometry of discrete groups", chapter 9. As soon as you have more than 6 vertices, you have a free choice for a pair of points that you cannot compensate by a global isometry (that operates triply transitively on the half-plane but no more). So you have uncountably many different subgroups that are of finite covolume and no two of them are conjugated.

I would guess that as soon as you have more than three cusps, the Teichmuller space associated to the surface is non trivial (a complex manifold with dimension > 0) whereas there are countably many (up to isometry) surfaces whose associated group is commensurable to $PSL_2(\mathbf{Z})$. So most discrete groups associated to a surface with sufficiently many cusps (4 probably) are not commensurable to $PSL_2(\mathbf{Z})$.

All discrete subgroups of $PSL_2(\mathbf{R})$ with finite covolume are obtained as follows. Choose a 2n-gone in the hyperbolic Poincare half-plane all of whose vertices are at infinity and such that opposite sides are identified through an isometry of the half-plane. The group generated by these isometries has the polygon as its fundamental domain, and since its vertices are at infinity, this fundamental domain has finite volume. All finite covolume discrete groups without elliptic elements are obtained this way. This is a particular case of Poincare theorem for surfaces, that shows that there is a correspondance between finitely generated subgroups of $PSL_2(\mathbf{R})$ and convex fundamental polygons in the half-plane. See Beardon, "The Geometry of Discrete Groups", chapter 9. As soon as you have more than 6 vertices, you have a free choice for a pair of points that you cannot compensate by a global isometry (that operates triply transitively on the half-plane but no more). So you have uncountably many different subgroups that are of finite covolume and no two of them are conjugated.