A space $X$ is filter compact iff every nonempty filter has at least one cluster point. I'll show that this is equivalent to the usual definition of compactness which I'll cover cover compactness. Annoyingly I couldn't find any proof online so I had to come up with the following proof myself.
Theorem. $X$ is filter compact $\iff$ $X$ is cover compact.
Proof. ($\impd$): Suppose $X$ is not filter compact. That means there is a nonempty filter $F$ such that every $x\in X$ has a nbhd $U_x$ which is disjoint from some $S_x\in F$. The $U_x$'s form an open cover of $X$ and assuming cover compactness, there is a subcover $\\{U_i\\}_{i\in I}$ indexed by the finite set $I$. Then, the finite intersection
$$\bigcap_{i\in I} S_i$$
is nonempty since $F$ is nonempty. But at the same time, this intersection is disjoint from all $U_i$'s, contradicting the claim that $\\{U_i\\}_{i\in I}$ covers $X$.
($\imp$): Suppose $X$ is not cover compact. So there is an open cover $\\{U_i\\}_{i\in I}$ with no finite subcover. To give an overview, I construct an increasing ordinal-indexed sequence of opens whose union is $X$, and then consider the filter generated by the complements.
By well-ordering the index set $I$, we can assume it is some ordinal $\gamma$. Then, for all $\beta\le\gamma$, define
$$V_\beta := \bigcup_{\alpha<\beta} U_\alpha.$$
Each $V_\beta$ is open in $X$, and they form an increasing sequence.
Now, let $\kappa$ be the least ordinal such that $V_\kappa=X$. I want $\kappa$ to satisfy two conditions:
The second condition is automatically satisfied because $\{U_i\}$ has no finite subcover. The first condition may not be true in general: for instance, if $X=(0,1]$ and the open cover consists of intervals $(1/n,3/4)$ and $(1/2,1]$, then one might have the sequence
$$\begin{matrix} \beta & 1 & 2 & 3 & \ldots & \omega & \omega+1 \\ V_\beta & (1/2,\,3/4) & (1/3,\,3/4) & (1/4,\,3/4) & \ldots & (0,\,3/4) & (0,\,1], \end{matrix}$$
where $\kappa=\omega+1$. Luckily, we can circumvent this issue: write $\kappa=\kappa'+n$ where $\kappa'$ is a limit ordinal and $n$ is finite, then form a new increasing sequence of opens by moving the last $n$ opens to the front:
$$V'_\beta := \left(\bigcup_{\kappa'<\alpha\le\kappa} U_\alpha\right) \cup \left(\bigcup_{\alpha<\beta} V_\alpha\right).$$
It can be seen that $\kappa'$ is the least ordinal satisfying $V'_{\kappa'}=X$. For instance, the sequence above is transformed into
$$\begin{matrix} \beta & 1 & 2 & 3 & \ldots & \omega \\ V_\beta & (1/2,\,1] & (1/3,\,1] & (1/4,\,1] & \ldots & (0,\,1]. \end{matrix}$$
Let $\{V_\beta\}$ be our increasing sequence of opens whose corresponding ordinal $\kappa$ satisfies the two conditions above. Now, define $F$ to be the filter generated by the complements $V_\beta^c$ where $\beta<\kappa$. Note that the complements form a nested decreasing sequence and none of them are empty by definition of $\kappa$, so they satisfy the finite intersection property and thus $F$ is nonempty.
Lastly, we show that $F$ has no cluster point. We have
$$X = V_\kappa = \bigcup_{\alpha<\kappa} V_\alpha,$$
where the second equality holds because $\kappa$ is a limit ordinal. For instance, if $\kappa=\omega+1$ then
$$\begin{align*} V_\kappa &= \left(\bigcup_{n\in\N} U_n\right) \cup U_\omega, \quad\text{whereas} \\ \bigcup_{\alpha<\kappa} V_\alpha &= \bigcup_{n\in\N} U_n. \end{align*}$$
Therefore every $x\in X$ is contained in some $V_\alpha$ where $\alpha<\kappa$, so $x$ is disjoint from $V_\alpha^c\in F$. But $V_\alpha^c$ is closed so some nbhd of $x$ is also disjoint from $V_\alpha^c$, therefore $x$ is not a cluster point of $F$. $\qed$