Start from a finite simplicial set $K$ which is homotopicaly equivalent to a Kan complex $X$.
Then by applying a finite number of pushout of outer horn inclusion to $K$, you can build homotopy equivalences $K \hookrightarrow K' \rightarrow X$ such that all the $1$-cells of $K'$ are "invertible" (in the sense that "for all $1$-cell $f$ there exists $2$-cells attesting homotopies $g \circ f => 1$ and $f \circ h => 1$ " see the "edit" below though ). $K'$ is still a finite simplicial set.
I claim that $K' \rightarrow X$ is now an equivalence in the Joyal model structure, which conclude the proof.
Indeed, as all the $1$-cells of $K'$ have homotopy inverses, the homotopy category of $K'$ (in the sense of the left adjoint to the nerve functor) is a groupoid.
So if I take $K' \hookrightarrow Y \rightarrow X$ a factorization as a Joyal trivial cofibration followed by a Joyal fibration, $Y$ is a quasi-category whose homotopy category is equivalent to the homotopy category of $K'$, hence is a groupoid, hence $Y$ is a Kan complex.
And $Y \rightarrow X$ is a homotopy equivalences between Kan complexes, hence it is a Joyal equivalence. So as announced, $K' \rightarrow X$ is a Joyal equivalence.
Edit: small correction and answering your comment. You are indeed right that it is not exactly possible to get what I said. What we need to do precisely is the following:
For each $1$-cell of $K$ you use a pushout by a $\Lambda^0  \hookrightarrow \Delta$ and one by a $\Lambda^2  \hookrightarrow \Delta$ to add a cells $g$ and $h$ with $2$-cells $f \circ g => 1$ and $h \circ f => 1$.
And you stop there, we don't add any new cells (no right inverse for $g$, or left inverse for $h$)
This is enough to ensure that the homotopy category of $K'$ is a groupoids: the original cells of $K$ , like $f$, will be invertible because they have both a left inverse and a right inverse, and the new cells $g$ and $h$ are invertible because they are either right or left inverse to an invertible cell.
As every arrow the homotopy category of $K'$ is a composite of $1$-cell of $K'$ they will all be invertible.