How can I see that there is no lattice in $G=\mathrm{PSL}_2( \mathbb{R})$ which contains $\Gamma_1=\mathrm{PSL}_2( \mathbb{Z})$ properly, or equivalently, that $X_1 =\mathrm{PSL}_2(\mathbb{Z}) \backslash \mathbb{H}$ is not a finite sheated nontrivial cover of a Riemann surface.

As a motivation, the eigenvalues of the Laplace Beltrami operator on $\mathrm{PSL}_2(\mathbb{Z}) \backslash \mathbb{H}$ is conjectured to have only eigenvalues of multiplicity one. So now, if there exists $\Gamma \supset \Gamma_1$, an eigenvalue on $X_1$ can be associated to an irreducibel representation $\pi$ of $\mathrm{Ind}_{\Gamma_1}^{\Gamma} $, since $Ind_{\Gamma}^{G}$ $ \mathrm{Ind}_{\Gamma_1}^{\Gamma} 1 $ and $\mathrm{Ind}_{G}^{\Gamma}1 $ are isomorphic, and the eigenvalues appear with multiplicty being square of the dimension of $\pi$ (eigenfunctions =matrix coefficients).

Hence, if there would exists such a thing with a higher dimensional representation, the conjecture would be wrong. That's why the intuition that something should be known here.

What about $G$ reductive (see comment of David Loeffler)!