Take any Dyck path of length $2n$ and observe the points bounded by it above the diagonal. This is precisely a coin stack of base $n$ since every layer has one less place to put coins than the layer below, and the different "peaks" of the Dyck path result from how many coins you've stacked in each layer.
Inversely, you can start with a coin stack, add a bottom layer of $n+1$ coins to it and draw its boundary. This will be a Dyck path as it can never go below the base (ie the diagonal) and it requires exactly $n$ moves of "going up" and $n$ moves of "going right". The last claim is pretty obvious if you imagine the path is from $(0,0)$ to $(n,n)$.
Here is how the bijection looks like for $n=4$. (added as a link as my reputation is insufficient to post images). the points on the path are colored blue, the points on the diagonal are crossed out, and the points inside the stack are colored black.