As $SL(2,\mathbb{C})$ is a double cover of the Lorentz group, is $SL(2,\mathbb{Z})$ a discrete subgroup of the Lorentz group?

As emphasized in the other answers, $SL(2,\mathbb{Z})$ is a discrete subgroup of the double cover of the Lorentz group (actually of its connected component), so $SL(2,\mathbb{Z})/\{\pm 1\}$ is a discrete subgroup of the Lorentz group itself. The purpose of this new answer is to say something about the geometric significance of this discrete subgroup.

To get some geometric insight, represent a vector with components $(t,x,y,z)$ as a $2\times 2$ matrix $$ X := \left(\begin{matrix} t+z & x+iy \\ x-iy & t-z \end{matrix}\right) $$ with $X^\dagger=X$ and $\text{det}(X)=t^2-x^2-y^2-z^2$, so that a Lorentz transformation can be expressed as $$ X \mapsto M X M^\dagger $$ with $M\in SL(2,\mathbb{C})$. The fact that $\pm M$ both give the same transformation of $X$ corresponds to the fact that $SL(2,\mathbb{C})$ is the double cover of the Lorentz group.

One thing to notice immediately is that the $y$ component of $X$ is invariant under every $M\in SL(2,\mathbb{Z})$. This follows from the fact that the real and imaginary parts of $X$ are separately self-contained under such a transformation, and from the fact that the reflection $y\to -y$ is excluded from the connected component. Therefore, after quotienting by $\{\pm 1\}$, the set of trasnformations with $M\in SL(2,\mathbb{Z})$ implements a subgroup of the Lorentz group of the three-dimensional spacetime with coordinates $t,x,z$.

Now, recall that the modular group $SL(2,\mathbb{Z})$ is generated by the two transformations $$ \left(\begin{matrix} 0 & 1 \\ -1 & 0 \end{matrix}\right) \hskip1cm \text{and} \hskip1cm \left(\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}\right). $$ The first of these implements the Lorentz transformation $$ \left[\begin{matrix} t \\ x \\ z \end{matrix}\right] \mapsto \left[\begin{matrix} t \\ -x \\ -z \end{matrix}\right], \tag{1} $$ and the second one implements the Lorentz transformation $$ \left[\begin{matrix} t \\ x \\ z \end{matrix}\right] \mapsto \left[\begin{matrix} (3t+2x-z)/2 \\ t+x-z \\ (t+2x+z)/2 \end{matrix}\right]. \tag{2} $$ Notice that the coefficients in this transformation are not all integers, although they are rational. As a check, we can confirm directly that this really is a Lorentz transformation: $$ \left(\frac{3t+2x-z}{2}\right)^2 - (t+x-z)^2 - \left(\frac{t+2x+z}{2}\right)^2 = t^2-x^2-z^2. $$ The geometric significance of (1) is obvious, but (2) is more difficult to visualize.