Here's a way to do it:

Recall that $C_n$ counts the number of lattice paths from $(0,0)$ to $(2n,0)$ taking only steps of the form $(1,\pm 1)$ that never goes below the $x$-axis; call this a Dyck path. Further, $$\frac{1}{\sqrt{1 - 4x}} = \sum \binom{2n}{n}x^k$$ which counts the total number of paths from $(0,0)$ to $(2n,0)$; call this a bridge. Also, $\binom{a+2n}{n}$ is the number of lattice paths (with the same step set) from $(0,0)$ to $(2n+a,a)$, since we have $a + 2n$ steps total with $n$ down steps (and thus $a + n$ up steps); call this an upward path.

Every upward path can be decomposed into:

and so on.

This provides a bijection from a single bridge with an $a$-vector of Dyck paths. Since the generating function for a single bridge with $a$-vector of Dyck paths is exactly the left-hand-side of your equality, it must equal the right-hand side.