Here's a variant on Joseph O'Rourke's construction that produces $2^{n/2}$ paths.  I don't have Joseph's knack for drawing pictures, so I'll describe it in words with an example.  (I'd be delighted if he would replace my words with a picture.)

Let's connect the origin $(0,0)$ to $(0,10)$.  Start by laying down a series of extremely long, extremely skinny, "blue" triangles that come in horizontally from the left to $(1,1)$, from the right to $(-1,3)$, from the left to $(1,5)$, from the right to $(-1,7)$, and from the left to $(1,9)$.  At this point there's just one shortest path, which zigzags from $(0,0$ to $(1,1)$ to $(-1,3)$ to $(1,5)$ to $(-1,7)$ to $(1,9)$ to $(0,10)$.  Now on each of these segments put one of O'Rourke's little yellow "splitters" -- in particular, say, at $(0,2)$, $(0,4)$, $(0,6)$, and $(0,8)$.

In general, if you use $k$ long blue triangle to create a zigzag path and put a yellow splitter on each of the $k+1$ segments of the zigzag, you get $2^{[(n+1)/2]}$ shortest paths, where $n=2k+1$.