Orbits of the projective special linear group on $\mathbf{Q} \cup \{\infty\}$  The group $\mathrm{PSL}_2(\mathbf{Q})$ of fractional linear transformations $x \mapsto (ax+b)/(cx+d)$ such that $a,b,c,d \in \mathbf{Q}$ and $ad-bc = 1$ acts on $\mathbf{Q} \cup \lbrace \infty\rbrace$. The weakest version of my question is the following:

Is there a short proof that no element of $\mathrm{PSL}_2(\mathbf{Q})$ acts transitively on $\mathbf{Q} \cup \lbrace \infty\rbrace$?

It seems obvious to me that no element of $\mathrm{PSL}_2(\mathbf{Q})$ can have finitely many orbits on $\mathbf{Q} \cup \lbrace\infty\rbrace$, but I have not been able to prove this.

Is this stronger claim true, and if so, does it have a short proof?

Some remarks and motivation: (1) The subgroup of $\mathrm{PSL}_2(\mathbf{Q})$ generated by the maps $x \mapsto x+1$ and $x \mapsto x/(x+1)$ acts transitively on $\mathbf{Q} \cup \lbrace \infty\rbrace$; these elements are used to generate the Calkin–Wilf tree of rational numbers.
(2) By conjugating by elements of $\mathrm{PGL}_2(\mathbf{Q})$ it suffices to look at maps of the form $g : x \mapsto 1/(b-x)$. If $|b| > 2$ then $g$ is conjugate in $\mathrm{PGL}_2(\mathbf{R})$ to a map of the form $x \mapsto \lambda x$. It follows that the iterates of $g$, starting at the initial value $x_0$ are of the form
$$ \frac{1}{b-\frac{1}{b-\ldots\frac{1}{b-x_0}}} $$
and converge to the continued fraction $1/(b-1/(b- \ldots ))$. The iterates of $g^{-1}$ behave similarly. So in this case each orbit of $\langle g \rangle \le \mathrm{PSL}_2(\mathbf{Q})$ has precisely two limit points in $\mathbf{R}$, namely the two roots of $x^2-bx+1$. Clearly this implies that $\langle g \rangle$ has infinitely many orbits on $\mathbf{Q} \cup \lbrace \infty\rbrace$. 
However if $|b| < 2$ then I have not been able to make this approach work.
(3) If $\mathbf{Q}$ is replaced with the finite field $\mathbf{F}_p$ then
is not too hard to show that no element of $\mathrm{PSL}_2(\mathbf{F}_p)$ acts transitively on $\mathbf{F}_p \cup \lbrace \infty \rbrace$: see Lemma 8.2 in this paper with John Britnell. This leads to a proof of the result for $\mathbf{Q}$ by reduction mod $p$.
 A: If the eigenvalues of $g$ are irrational, then they generate a quadratic extension field $K$ of $\mathbb Q$. Then the desired statement is that in the group $K^\star/\mathbb Q^\star$ the subgroup generated by a single element does not have finite index. But $K^\star/\mathbb Q^\star$ is not finitely generated, since there are infinitely many primes that split in $K$.
Edit: In more detail, you then have a basis for $K$ such that the map $K\to K$ given by multiplication by an eigenvalue corresponds to the matrix $g$. The cases with rational eigenvalues are easy.
A: If your matrix is scalar it acts trivially on the projective line, so assume it's not scalar; hence it's conjugate in $\text{GL}_2(\mathbf{Q})$ to a companion matrix $A=\pmatrix{0 & -1 \cr 1 & z}$. 
If $z$ is integer with $|z|\ge 3$ then the dynamics of $A$ on the real projective line $\mathbf{P}^1(\mathbf{R})$ has a repelling point $a_-$ and a attractive point $a_+$; in particular, if $K$ is an infinite compact subset of $\mathbf{P}^1(\mathbf{R})\smallsetminus (a_-,a_+)$, then its intersection with each $A$-orbit $\{A^nx:n\in\mathbf{Z}\}$ is finite and hence $K$ intersects infinitely many $A$-orbits. Since $K$ can be chosen to be contained in $\mathbf{P}^1(\mathbf{Q})$, it follows that $A$ admits infinitely many orbits on $\mathbf{P}^1(\mathbf{Q})$. If $|z|=2$ a similar argument works (with $a_+=a_-$). If $|z|<2$ then $A$ has finite order.
Assume now $z$ is not an integer. Let $p$ be a prime divisor of the denominator of $z$. Let $|z|_p$ be the norm of $z$ in $\mathbf{Q}_p$. An easy calculation shows that the function $x\mapsto -1/(x+z)$ maps $\mathbf{Z}_p$ into itself and is $(1/|z|_p^2)$-contracting on 
$\mathbf{Z}_p$,
so it admits a unique fixed point which I denote by $a_+$. Thus $A$ admits a north-south dynamics on $\mathbf{P}^1({\mathbf{Q}_p})$, with repelling and attracting points $a_-$ and $a_+$. So the same argument as in the real case, using density of $\mathbf{P}^1({\mathbf{Q}})$ in $\mathbf{P}^1({\mathbf{Q}_p})$ shows 
that $A$ has infinitely many orbits. I assumed (as you do) that $A$ has determinant 1 but the argument carries over with minor modification for arbitrary $A$.
A: The stronger claim is indeed true. Assume $g \in PSL_2(\mathbb{Q})$ has infinite order, equivalently $|trace(g)| \ge 2$. If necessary replace $g$ by its square to make the trace positive. The action of $g$ on the circle $\mathbb{R} \cup {\infty}$ comes in one of two flavors, depending on whether $trace(g) = 2$ or $>2$. When $trace(g)=2$ then $g$ has a single fixed point in $\mathbb{R} \cup {\infty}$ and acts properly discontinuously on the complementary line (topologically conjugate to a translation acting on the real number line). When $trace(g)>2$ then $g$ has two fixed points and acts properly discontinuously on each of the two complementary lines. In either case, since $\mathbb{Q} \cup {\infty}$ is dense in each such line, $g$ has infinitely many orbits.
Edit: Yves' comment shows that there is a hole in this approach due to infinite order elliptic elements in $SL(2,\mathbb{Q})$. Not sure whether it can be patched it up over $\mathbb{R}$, but Yves' answer, which for the case $trace(g)<2$ uses north-south dynamics over $\mathbb{Q}_p$, does the trick very nicely.
A: Consider the matrix $A=\begin{pmatrix}a&b \cr c&d\end{pmatrix}$ in ${\rm SL}(2,\mathbb Z)$. Then you're asking if setting $\begin{pmatrix}p_n\cr q_n\end{pmatrix}=A^n\begin{pmatrix}p\cr q\end{pmatrix}$ all of ${\mathbf P}^2(\mathbb Q)$. To see that it's not, notice that there are a handful of cases: 
$A$ is elliptic if it has a pair of eigenvalues on the unit circle. These are necessarily roots of unity. In this case, $A$ has finite order, and so it's obvious that there are infinitely many orbits. 
$A$ is parabolic if it has a repeated root of $\pm 1$. In this case, $A^n\begin{pmatrix}p\cr q\end{pmatrix}$ converges in ${\mathbf P}^2(\mathbb Q)$ to a fixed direction (the generalized eigenvector) unless $\begin{pmatrix}p\cr q\end{pmatrix}$ was in the direction of the eigenvector
Similarly if $A$ is hyperbolic, there are two possible limit directions.
This rules out the transitivity you were looking for.
