The Poincare recurrence theorem states that for any $f$-invariant probability measure $\mu$ on $X$ and any set $A$ with $\mu(A) > 0$, $\mu$-a.e. point $x \in A$ returns to $A$, i.e. there exists $n > 0$ s.t. $f^n(x) \in A$.

If $X$ is second countable and Hausdorff and $\mu$ is Borel, then you can remove the sets of zero measure from all elements in a countable neighborhood basis to immediately show that $\mu$-a.e. point is recurrent. And by the Bogoliuboff-Kryloff theorem, there is always at least one such measure $\mu$; take a weak-* limit point of averages of the form $\frac{1}{n} \sum_{i=0}^{n-1} f^i \nu$ for an arbitrary (not necessarily invariant) measure $\nu$.

So the only case where there could be only a single recurrent point $x$ is if $\delta_x$ is the only $f$-invariant Borel measure, which is equivalent to saying that $\frac{1}{n} \sum_{i=0}^{n-1} f^i \delta_y$ converges weak-* to
$\delta_x$ for all $y \in X$. This, in turn, should be equivalent to saying that for any $y \in X$, the orbit $(f^n(y))$ converges to $x$ except for a set of $n$ of zero density.

So, as long as it is not the case that every orbit in your system converges to the same limit except for a set of zero density, you'll have at least two recurrent points.