Let $S_k$ be the set of permutations on $\{1,\dots,k\}$.
Theorem: For all $k\ge 2$, $\Delta(k,m)$ is equal to $(-1)^{k-m}$ times the number of permutations in $S_k$ with exactly $m$ cycles and where $1$ and $2$ are in the same cycle.
Proof: Let $S_k^{m}$ be the set of permutations in $S_k$ with at most $m$ cycles. Since a permutation in $S_k$ with $i$ cycles has a sign of $(-1)^{k-i}$, the sum $\Delta(k,m)$ is equal to the sum of the signs of all of the permutations in $S_k^{m}$. To simplify this summation, we shall partition $S_k^m$ into pairs and singletons, where the two permutations in each pair have opposite signs. These pairs cancel each other out in the summation, so we can sum over the singletons alone.
For each $\pi \in S_k^m$, we will pair $\pi$ with $(1\;\;2)\pi$, except in the case where $(1\;\;2)\pi$ has more than $m$ cycles. Note that $(1\,\,\,2)\pi$ has the opposite sign of $\pi$. This causes all permutations in $S_k^m$ to be paired up, where the only exceptions are the permutations with exactly $m$ cycles, such that $1$ and $2$ are in the same cycle. Since all of the unpaired permutations have a sign of $(-1)^{k-m}$, the entire sum equals $(-1)^{k-m}$ times the number of such permutations.
$\tag*{$\square$}$
As a corollary, we have
$$
\left|\Delta(k,m)\right| \le {k\brack m}
$$
There may be some combinatorial argument which allows you to get a more precise count on the number of permutations with $m$ cycles such that $1$ and $2$ are in the same cycle, and thus to get a better bound than above. However, the above is the best bound I can give.
