It seems that the number of such topologies is $2^{\mathfrak c}$. Such (huge) number of connected Hausdorff topologies can be constructed by a suitable modification of the Bing's construction of a connected Hausdorff space.

First we introduce the necessary definitions. A subset $U$ of the real line is called *regular open* if $U$ coincides with the interior of the closure of $U$ in $\mathbb R$.

A filter $\mathcal F$ on $\mathbb R$ will be called

$\bullet$ (*regular*) *open* if $\mathcal F$ has a base consisting of (regular) open subsets of $\mathbb R$;

$\bullet$ *vanishing* if each neighborhood of zero in $\mathbb R$ belongs to $\mathcal F$.

It can be shown that the set of regular open vanishing filters on $\mathbb R$ has cardinality $2^{\mathfrak c}$.

For any open vanishing filter $\mathcal F$ we define a countable connected Hausdorff space $X_{\mathcal F}$ as follows. On the countable set $X=\{(x,y)\in\mathbb Q\times\mathbb Q:y\ge 0\}$ consider the topology $\tau_{\mathcal F}$ in which a neighborhood base at a point $(x,y)\in X$ consists of the sets
$$\Delta_U=\{(x,y)\}\cup\{(t,0)\in X:t\in (x-\sqrt{3}y+U)\cup(x+\sqrt{3}y+U)\}$$where $U\in\mathcal F$.

It can be shown that the topological space $X_{\mathcal F}$ is Hausdorff and the closures of any non-empty open sets in $X_{\mathcal F}$ have a common point, which implies that $X_{\mathcal F}$ is connected.

It can be shown that for distinct regular open vanishing filters $\mathcal U,\mathcal V$ the topologies $\tau_{\mathcal U}$, $\tau_{\mathcal V}$ on the set $X$ are distinct, which implies that on the countable set $X$ there exist $2^{\mathfrak c}$ Hausdorff connected topologies. Since the cardinality of bijections of the set $X$ is continuum, the number of pairwise non-homeomorphic connected Hausdorff topologies on $X$ remains equal to $2^{\mathfrak c}$.