Yes there is if the finite number of pieces are large enough. But to answer your question I want black to win (because I hate it when problems most often require white to win). Let's consider the case for $N\times N$ boards and extrapolate it and then I will show how this is done for infinite boards. For a usual $8\times 8$ board there is no argument on the fact that we have mates in $1,2,\dots,7$ (and Im sure even more). I claim that for an $N\times N$ board we can have mates in $1,2,\dots,N-1$, and then I show how this is done for an infinite board.

Consider this classical mate in 7 (white to move, black mates in 6 plies, white moves maximum 7 plies) position (actually its mate in 8 if a queenside castling was allowed and we had black rook in a8 and king in e8, its funny position I always show people who never considered castling when solving such problems):

![alt text][1]

Now you can reproduce the same position for an $N\times N$ board and get a mate by $N-1$ for any $N>8$. To make this work for an infinite board just surround an infinite board by black's pawn to "create" a bounded $N\times N$ board. So to make an $8\times 8$ board like the one in the diagram below. Just surround the $10\times 10$ area by Black's pawn. White's king cannot move away from the "boundary" because of the pawns. So in this way we see that we get mate in $7,8,9,\dots$ for an infinite board. For $1,2,\dots,6$ moves to mate, we do the same by only creating an $8\times 8$ board but hopefully in such a way that the "boundary" cannot be taken or moved by white (I don't think its difficult to provide the particular examples here). 

  [1]: http://i.harepix.com/i/389837748.jpg

My instincts tell me that this particular example can be done for any ordinal as well.