I can prove that 12 is the best you can do:
First, let's agree to call a move of form $(a, b)$ a move that is $a$ vertical steps and $b$ horizontal steps, or vice-versa. So, for example, a standard knight move is $(2, 1)$.
Step 1: Notice that a speedy knight can make all moves
$$(2, 0), (2, 2), (3, 1), (3, 3), (4, 0), (4, 2),$$
anywhere on the board. Also, a speedy knight can make any move
unless one endpoint of the move is in the corner -- I'll call this the corner exception.
In other words, if a move $(a, b)$ has $a$ and $b$ at most $4$, and same parity, then it is legal unless it is $(2, 2)$, $(4, 4)$ or the corner exception.
Step 2: By symmetry it's enough to show that there can be at most $3$ knights on light-colored squares in the bottom half of the board. (Notice light- and dark- colored squares don't interact.) Let's agree to number the columns 1 through 8.
Step 3: Consider two knights on light-colored squares in the bottom half of the board. The vertical distance between them is at most 3. I claim the horizontal distance between them can't be 0 (same column), 1 (adjacent columns), 3, or 4, except for the corner exception. For example if the horizontal distance was 3 then it would either be a $(3, 1)$-move or a $(3, 3)$-move between them, and either way they are attacking each other. The other cases are easy to check.
Step 4: Note that the corner exception only occurs in columns 7 and 8. (Remember, we're looking at light-colored squares on the lower half of the board.)
Step 5: We're almost done! Notice there can be at most:
-one knight in columns 1, 2, 5
-one knight in columns 3, 6, 7
-one knight in columns 4, 8.
Hence, at most three knights total. The proof is complete.