With the problem as stated, the answer is $n-k+1$. Take the uniform matroid of rank $1$ on $n-k+1$ elements and direct sum with $k-1$ co-loops. (Geometrically, take the standard basis $e_1$, $e_2$, ..., $e_k$ of $\mathbb{R}^k$ and duplicate the first basis element $n-k+1$ times.)


To see that this is optimal, suppose for the sake of contradiction that we could achieve $n-k$ bases: $B_1$, $B_2$, ..., $B_{n-k}$. Consider the exchange graph, where $B_i$ and $B_j$ are connected by an edge if $\#(B_i \setminus B_j)= \#(B_j \setminus B_i) = 1$. It is known to be connected. Reorder the $B_i$ such that the induced subgraph on $B_1$, $B_2$, ..., $B_r$ is connected. Then, for $2 \leq r \leq n-k$, there is at most one element of $B_r$ not in $\bigcup_{1 \leq i<r} B_i$. We deduce that $\# \bigcup_{1 \leq i \leq n-k} B_i \leq \# B_1 + (n-k-1) = n-1$. So there is some element not in $\bigcup B_i$, and this element is not independent.

As I commented above, I think it would be more natural to impose that $M$ has neither loops nor co-loops. The best I can find for that problem $k(n-k)$, by using $U(k-1,k) \oplus U(1, n-k)$, where $U(r,m)$ is the uniform matroid of rank $r$ on $m$ elements. If you ask for the matroid to be connected, I can achieve $k(n-k)+1$ by starting with $U(k,k+1)$ and replacing one element with $n-k$ parallel copies. I would guess these are optimal.