One possibility for closed maps is
$$f \text{ closed}\iff \bigg( f(\mathcal{F}) \to y \implies \mathcal{F} \to f^{-1}(\{y\}) \bigg), $$
or equivalently
$$f \text{ closed}\iff \bigg( f(\mathcal{F}) \text{ clusters at } y \implies \mathcal{F} \text{ clusters at } f^{-1}(\{y\}) \bigg), \tag{1}$$
where the notion of a filter $\mathcal{F}$ clustering at a set $S$ means that every $F$ in $\mathcal{F}$ meets every neighborhood of $S$, and similarly for convergence. Note that we can allow $f^{-1}(\{y\})$ to be empty, if we consider "clustering at the empty set" to be a vacuous condition.
Analogous statements apply if we replace filters by nets.
Note if $f$ happens to be proper, then $f^{-1}(\{y\})$ is compact, and "$\mathcal{F}$ clusters at $f^{-1}(\{y\})$" is equivalent to "$\mathcal{F}$ clusters at some point in $f^{-1}(\{y\})$". Interestingly, the condition "$f(\mathcal{F})$ clusters at $y \implies \mathcal{F}$ clusters at some $x \in f^{-1}(\{y\})$" is equivalent to $f$ being universally closed (see Bourbaki Topology section 10.2).
Another, more esoteric possibility for closed maps is the following: consider the functor $f_\forall :\mathcal{P}(X) \to \mathcal{P}(Y)$ given by
$$f_\forall (S) := \{ y \in Y: f(x)=y \implies x \in S\},$$
and denote by $f_\exists$ the normal direct image. Then $f_\exists$ takes closed sets to closed sets iff $f_\forall$ takes open sets to open sets, since $f_\exists (S^c) = f_\forall(S)^c$. So assuming the point in the original post is correct, a map is closed iff $\langle f_\forall (\mathcal{N}_x) \rangle \subseteq \langle \mathcal{N}_{f(x)} \rangle$ for all $x$, where $\langle \mathcal{B} \rangle$ represents the filter generated by filter base $\mathcal{B}$.