Two topologies are equal if they have the same filter convergence

This is true.

To see why just recall that $x\in \overline{A} \iff$ there is an ultrafilter $\mathcal{F}\rightarrow x$ with $A\in \mathcal{F}$. And show the closures of any subset must be identical in both cases.

For any topological space $X$:

$O$ is open iff for all $x \in O$ and for every ultrafilter $\mathcal{F}$ that converges to $x$, we have $O \in \mathcal{F}$.

Left to right is clear, as $O$ with $x \in O$ is a neighbourhood of $x$, so $O$ must be in any ultrafilter that converges to $x$.

Right to left: suppose $O$ satisfies the condition, but is not open, so some point $p \in O$ is not an interior point of $O$. This means that all neighbourhoods of $x$ intersect $X\setminus O$, and so $\{X\setminus O\} \cup \mathcal{N}_x$ forms a filter base so there is some ultrafilter $\mathcal{F}$ that contains $\mathcal{N}_x$ and $X\setminus O$.
Contradiction, as $\mathcal{F} \to x$ so $O \in \mathcal{F}$ by the condition, but then this ultrafilter contains disjoint sets.

So it follows that if the ultrafilter convergence is the same, the open sets will be the same too.

If $\tau_1\neq \tau_2$, there is a set which is open in one topology but not the other. Suppose (swapping the roles of $\tau_1$ and $\tau_2$ if necessary) $U\in \tau_1$ but $U\notin \tau_2$. Since $U$ is not open in $\tau_2$, there is some point $x\in U$ such that no $\tau_2$-neighborhood of $x$ is contained in $U$. It follows that the set $\{V\mid x\in V\text{ and }V\in \tau_2\}\cup \{X\setminus U\}$ has the finite intersection property, so it extends to an ultrafilter $\mathcal{F}$ on $X$.

Now $\mathcal{F}\to x$ in $\tau_2$, since $\mathcal{F}$ contains all the $\tau_2$-open neighborhoods of $x$, but $\mathcal{F}\not\to x$ in $\tau_1$, since $\mathcal{F}$ does not contain $U$ (which is a $\tau_1$-open neighborhood of $x$).