I will decompose the problem of finding facets into 2 parts. First, let us obtain an irredundant representation of the system of inequalities. In the second part, we will identify implicit equalities and eliminate them as candidates for facets. Also thanks to @RobPratt for identifying a deficiency with my earlier answer.
Part 1: Let us assume we are given a system of inequalities (say $Ax \leq b \iff \lbrace{a_i^\top x \leq b_i \rbrace}_{i=1}^{L}$). Suppose we wanted to know if $a_1^\top x \leq b_1$ is redundant or not, we can compute the optimal value (OPT) of following linear program:
\begin{align}
OPT = \underset{x\in \mathbb{R}^{n}}{\max{}} &\,a_1^\top x \\
\mbox{s.t.} &\, a_i^\top x\leq b_i, i\in [2, L] \tag{1}
\end{align}
If $OPT \leq b_1$, then we know that the inequality $a_1^\top x \leq b_1$ is redundant (and so not a facet) since every point that satisfies the set of inequalities $\lbrace{ a_i^\top x\leq b_i \rbrace}_{i=2}^L$ already obey the inequality $a_1^\top x \leq b_1$. At this stage you must remove the inequality $a_1^\top x \leq b_1$ from the system $Ax \leq b$ to obtain an irredundant represenatation. If on the other hand $OPT > b_1$, then $a_1^\top x \leq b_1$ may be a candidate for a facet of $Ax\leq b$ and so you don't eliminate that inequality. We can repeat this process for the remaining $L-1$ inequalities one by one, to obtain the final set of irredundant inequalities.
Part 2
For the second part, we will assume that $Ax \leq b \iff \lbrace{a_i^\top x \leq b_i \rbrace}_{k=1}^{m}$ does not contain any redundant inequalities. Further denote $P = \lbrace{ x \in \mathbb{R}^n| Ax \leq b\rbrace}$. It may be the case that every $x \in P$ satisfies $a_i^\top x = b_i$ for some $i \in [1, m]$. If the condition holds then the inequality $a_i^\top x \leq b_i$ is an implicit equality and not a facet of $P$. To identify whether inequality is implicit equality or not, I recommend looking at the procedure described in the first answer here(https://cs.stackexchange.com/questions/128505/how-to-calculate-the-dimension-of-a-convex-polyhedron). For convenience I am reposting the relevant bit here. Suppose:
$$\max \{a_i^\top x \mid Ax\leq b\} = \min \{a_i^\top x \mid Ax\leq b\} = b_i,$$
then $a_i^\top x \leq b_i$ is an implicit equality. Whichever inequalities in $Ax \leq b$ are not implicit equalities are the facets.