Thinking about it some more, I believe all the ingredients are there to prove the minimum on an infinite board is 23 moves, or 24 positions.

If the termini are adjacent, we must reach 17 positions from the start. 12 positions are in two 6-cycles, so clearly we need 12 moves to reach them plus 1 move to get from one cycle to the other. There are four more positions that require 2 moves each to reach. (Or one could be reached in 1 move from the start position, but then you need 2 moves to reach the first 6-cycle.) 1 more move is required to reach the end. So we need at least 12+1+2×4+1=22 moves. But the termini are on opposite color squares, so the number of moves must be odd, so 23 is the minimum.

If the termini are a knight’s move apart, we must reach 15 positions from the start. 12 positions are in six 2-cycles, so we need 12 moves to reach them plus 5 moves to get between cycles. There are two more positions that require 2 moves each to reach. 1 more move is required to reach the end. So we need at least 12+5+2×2+1=22 moves. Again 23 is the minimum.

For any other separation of the termini, more moves are needed.∎