Suppose, without loss of generality, $m \le n$. If $m = 2$, Arnold has winning strategy if and only if $n$ is not a multiple of $4$; for $m \ge 3$, Arnold has winning strategy if and only if $m$ and $n$ are both odd.

Suppose $m = 2$. If $4$ does not divide $n$, Arnold can win placing the knight on the first column if $n = 4k + 1$ and on the second column if $n = 4k + r$, $r = 2$ or $r = 3$. The knight should always be moved two columns ahead on each move, allowing exactly $2k$ moves.
Now consider $n = 4k$. Divide the table into $2 \times 4$ subtables, and pair the cells in such a way that it's always possible to move the knight between cells from each pair:

Since the whole table is divided into pairs, Bernold can always move, no matter where Arnold places the knight: it suffices to move the knight to the other cell in the pair. When Arnold plays, he will move the knight to a cell from another pair, and Bernold can repeat this strategy. So if $n = 4k$ Bernold has winning strategy.
This finishes the case $m = 2$. The case $m \ge 3$ follows in a similar fashion, dividing the table into smaller subtables:

Those tables prove that Bernold has winning strategy for tables $3 \times n$, $n$ even and Arnold has winning strategy for tables $3 \times n$, $n$ odd: divide the table into one $3 \times 3$ or $3 \times 5$ subtable and several $3 \times 4$ subtables; it suffices to place the knight on the cell marked with an $A$ and then use the aforementioned Bernold's strategy.
The case $m = 4$ can be verified using several $4 \times 2$ subtables if $n$ is even and one $4 \times 3$ subtable and several $4 \times 2$ subtables if $n$ is odd. This also proves that if either Arnold or Bernold has winning strategy for a $m \times n$ table then he also has winning strategy for a $(m+4) \times n$ table, $m \ge 3$. So it suffices to solve the problem for $m \in \{3, 4, 5, 6\}$.
The case $m = 5$ can be solved using the following subtables:

Notice that Bernold has winning strategy for a $5 \times 4$ table and Arnold has winning strategy for a $5 \times 3$ table. Thus, in the case $5 \times n$, $n$ even, we merge $5 \times 4$ subtables if $n$ is a multiple of $4$ and one $5 \times 6$ subtable and $5 \times 4$ subtables if $n = 4k + 2$; if $n$ is odd, we merge several $5 \times 4$ subtables to a $5 \times 3$ or $5 \times 5$ subtable, if $n = 4k + 3$ or $n = 4k + 1$, respectively.
The case $6 \times n$ follows directly from the case $3 \times n$ if $n$ is even (merge two $3 \times n$ tables) and merging $6 \times 4$ subtables to a $6 \times 3$ or $6 \times 5$ subtable, if $n = 4k + 3$ or $n = 4k + 1$, respectively. All the cases are covered.