Sorry to be coming to the conversation so late (over 3 years late!), but I'm new to MO and just saw it. I've done some research on this very topic, so I couldn't resist sharing.
Let me start by saying that I don't know a complete answer to the question as stated. By I do have a partial answer, some other results that are relevant to the question, and an answer to the question of Doyle and Hocking mentioned in Michal's answer. It's all too long for a comment, so here we go . . .
The following theorem answers the Doyle and Hocking question, and also tells us that any example answering Ramiro's question affirmatively will be at least a little bit exotic.
Theorem: Suppose $X$ is a sequential space. Then either $br(X) = 1$ or $br(X) \geq 2^\mathfrak{c}$.
Proof: Let $Y$ be a space bijectively related to $X$, and let $f: Y \to X$ be a continuous bijection. Without loss of generality, we may assume that $Y$ and $X$ are actually just two topologies on the same set $A$ (because $f$ is a bijection), and $X$ is a refinement of $Y$ (because $f$ is continuous). Let us assume that $Y$ is strictly finer than $X$ (this must be true, in particular, if $Y$ is not homeomorphic to $X$).
Since $X$ is sequential and $Y$ is strictly finer than $X$, there is some sequence $\langle x_n \rangle$ such that $x_n \to x$ in $X$ but $x_n \not\to x$ in $Y$.
In $Y$, since $x_n \not\to x$, there is an infinite $S \subseteq \{x_n : n \in \omega\}$ such that $x \notin \overline{S}$. Thus (passing to a subsequence if necessary) we may assume that $x_n \to x$ in $X$ but $x \notin \overline{\{x_n:n \in \omega\}}$ in $Y$.
Let $\mathcal F$ be any free filter on $\omega$. Define the topology $Z_{\mathcal F}$ on $A$ as follows (we will define the space by defining its closure operator). If $B$ is any subset of $A$, then $\overline{B}$ is the same in $Z_{\mathcal F}$ as in $X$, except possibly for the point $x$. Then we put $x \in \overline{B}$ if and only if either $x \in B$, $x \in \overline{B \setminus \{x_n:n \in \omega\}}$, or $\{n : x_n \in B\} \in \mathcal F^+$.
It's easy to check that this defines a topology on $A$, and that this new topology is finer than $X$ and coarser than $Y$. This means that the identity on $A$ gives continuous bijections $Y \to Z_{\mathcal F} \to X$. So $Z_{\mathcal F}$ is bijectively related to $X$ for any filter $\mathcal F$.
Now let's check that we get $2^\mathfrak{c}$ non-homeomorphic spaces this way. Consider the following topological invariant:
$$Filt(Z) = \{\mathcal F : \mathcal F \text{ is a filter and, for some countable set } \{x_n: n \in \omega\},\text{ and some } x, \ \mathcal F = \{\{n : x_n \in U\}: U \text{ open and } x \in U\}\}$$
In other words, $Filt(Z)$ is just a list of all the filters that describe the convergence of some countable set to some point. Since $X$ is sequential, $Filt(X)$ is just the filter of cofinite sets. Given how we've defined $Z_{\mathcal F}$, $Filt(Z_{\mathcal F})$ is equal to either all isomorphs of $\mathcal F$ (if $X$ has one non-isolated point) or the filter of cofinite sets plus all isomorphs of $\mathcal F$ (if $X$ has more than one non-isolated point).
By an "isomorph" of $\mathcal F$ I mean every filter that can be obtained from $\mathcal F$ by a permutation of $\omega$. This class has size $\mathfrak{c}$ for any fixed $\mathcal F$. But the number of filters on $\omega$ is $2^\mathfrak{c}$, so $Filt(Z_{\mathcal F})$ is different for $2^\mathfrak{c}$ different choices of $\mathcal F$. Since it's a topological invariant, we have lots of different spaces.
QED.
Next, let's observe that for every infinite $\kappa$ there is some $X$ with $br(X) = \kappa$.
Fix two ultrafilters on $\omega$, say $p$ and $q$. Let $X_p = \omega \cup \{*\}$ be the space obtained by making every point of $\omega$ isolated, and making the neighborhoods of $*$ take the form $A \cup \{*\}$ for $A \in p$. Define $X_q$ and $X_{p \cap q}$ similarly (note that $p \cap q$ is a filter, so this definition still makes sense).
Let $\mathcal X$ be the topological space obtained by taking the disjoint sum of $\aleph_\kappa$ copies of $X_p$, $\aleph_\kappa$ copies of $X_q$, $\aleph_\kappa$ copies of $X_{p \cap q}$, and $\aleph_\kappa$ singletons.
As in the proof of the previous theorem, if $\mathcal Y$ is bijectively related to $\mathcal X$, then we may assume that $\mathcal Y$ is just a refinement of $\mathcal X$. By refining $\mathcal X$, we can obtain any space of the following form (for any $\mu,\lambda,\theta \leq \kappa$): a disjoint sum of $\mu$ copies of $X_p$, $\lambda$ copies of $X_q$, $\theta$ copies of $X_{p \cap q}$, and $\aleph_\kappa$ isolated points.
Using the fact that the only extensions of the filter $p \cap q$ are $p$ and $q$, it is not too hard to check that these are the only spaces possible.
Now suppose we also have $\theta = \aleph_\kappa$. There are $\kappa$ possibilities for choosing $\mu$ and $\lambda$, yielding $\kappa$ spaces of this form. But now we may refine again to get $\mathcal X$ back (leave $\aleph_\kappa$ of the $X_{p \cap q}$ alone, refine $\aleph_\kappa$ of them to $X_p$, and refine another $\aleph_\kappa$ of them to $X_q$).
Now I'll give a really nice example of a space $X$ where $br(X) \geq \mathfrak{c}$, namely the Baire space.
Unfortunately, I don't know a really slick proof of this assertion. I'll have to refer you to a paper of mine for details. Below I'll list a few related results that are also relevant to your question. For most of the results about non-separable spaces, see my other paper, joint with Arnie Miller.
I studied the "bijectively related" relation on (nonempty) perfect completely ultrametrizable spaces (henceforth, PCU spaces).
[Sidebar: At first this may seem like a weirdly specific class to look at. The following folklore result might clear things up a bit: $X$ is a PCU space if and only if there is some pruned, perfect tree whose end space is homeomorphic to $X$. So you can think of these as "tree spaces." The original intent of the two papers above was to explore how the structure of the trees can be leveraged to prove topological results.]
The separable PCU spaces are precisely the perfect, zero-dimensional Polish spaces (see Chapter 2 of Kechris's book for more on this). For this class, we have a pretty good idea of what your relation looks like:
Theorem: The separable PCU spaces are partitioned into exactly three equivalence classes by the relation described in your question: the class of the Cantor space, and the class of the Cantor space minus a point, and the class of the Baire space.
The latter two classes each have $\mathfrak{c}$ members. This result isn't in my paper, but it's a good exercise.
Once you move to non-separable PCU spaces, it is consistent that the situation stays "nice" like it is for Polish spaces. For example,
Theorem: If CH holds, then there are exactly four equivalence classes of PCU spaces of size $\mathfrak{c}$.
More generally,
Theorem: It is consistent with ZFC that $\mathfrak{c} = \aleph_n$ and that there are exactly $n+3$ equivalence classes of size-$\mathfrak{c}$ PCU spaces.
However, if $\mathfrak{c} = \aleph_2$ then the number of equivalence classes is independent of ZFC: it can be $5$ by the above theorem, but it is more if MA holds (I don't know exactly how many).
For cardinals below $\aleph_\omega$, there is a strong connection between these equivalence classes and the question of whether you can partition the Cantor space into $\kappa$ closed sets for $\kappa < \mathfrak{c}$.
I don't know much about what happens after $\aleph_\omega$, and I consider it a very interesting problem.
Two other papers of mine you might want to look at: this one deals with what happens when you refine a topology by just a little bit, and this one (joint with Chris Good, Robin Knight, and Dave McIntyre) deals with finite intervals in the lattice of topologies. (Notice that, by our arguments above, $X$ and $Y$ are bijectively related if and only if there is an interval in the lattice of topologies, with the top and bottom spaces both homeomorphic to $X$, and some space in between homeomorphic to $Y$. But then everything in the interval is also bijectively related to $X$ and $Y$. So one strategy for answering Romiro's question could be to find a finite interval like this).
