Consider three cases:

*Exactly one of the numbers $n$, $m$ is odd.* Color the squares like on a chessboard. Every move changes the color of the square where the knight is, whence after an odd number of moves, the knight is on a square of the opposite color. Thus the knight cannot be on the initial square after $2019$ moves.

*Both numbers $n$ and $m$ are odd.* Color the horizontal lines of the board alternately black and white. Again, every move changes the color of the square where the knight is. Hence, similarly to the previous case, the knight cannot be on the initial square after $2019$ moves.

*Both numbers $n$ and $m$ are even.* Let $n = 2^k p$, $m = 2^l q$ where $p$ and $q$ are odd. W.l.o.g., assume that $k \le l$ and the knight starts from square $(0,0)$. Obviously, the knight only visits squares with coordinates of the form $(2^k u, 2^l v)$ since $2^k$ divides the length of the step in either direction. Suppose that the knight is on the initial square after $2019$ moves. Shortening all moves $2^k$ times while retaining their directions, we obtain a route of a $p$-$2^{l-k}q$-knight which in $2019$ moves returns to the initial square. But $p$ is odd which means that such route does not exist by previous cases. The contradiction shows that the knight cannot be on the initial square after $2019$ moves.