Pairs of matroids $(M_1, M_2)$ with $\mathcal C(M_1) = \mathcal B(M_2)$

Yes there are more pairs like that. We know we can specify a matroid by its set of circuits and bases.

So now let $$B=C=\{X\subset\{1,2,3\}| |X|=2\}$$ Now consider the matroids $$M_1=(\{1,2,3,4\},\text{bases}=B)$$ $$M_2=(\{1,2,3,4\},\text{circuits}=C)$$ So $M_1$ and $M_2$ both have underlying set $\{1,2,3,4\}$, and there bases resp circuits specified as above. These will not be uniform matroids, but by definition have the required property.

Claim: Given two loop-free matroids $M_1$ and $M_2$, $\mathcal{B}(M_{1}) =\mathcal{C}(M_2)$ if and only if $M_1$ and $M_2$ are both partition matroids with $M_1 = \bigoplus_{i = 1} ^nU_{r_i,n_i}$ and $M_2 =\bigoplus_{i = 1} ^n U_{r_i-1,n_i}$.

Since bases (respectively, circuits) of a direct sum of matroids are just the disjoint union of bases (respectively, circuits) of the summands, it's enough to prove the above claim in the case when $M_2$ is connected.

Claim: Let $M_2$ be connected. Then $\mathcal{B}(M_1)=\mathcal{C}(M_2)$ if and only if there exist $1<r<n$ such that $M_1 = U_{r,n}$ and $M_2 = U_{r-1,n}$.

The "if" direction follows directly from the definition of uniform matroids. To prove the "only if" direction first let $r$ be the rank of $M_1$. Then every circuit of $M_2$ has cardinality $r$ and so by connectivity $M_2$ has rank $r-1$. So by Exercise 1.3.4 in Oxley's Matroid Theory, $M_2$ is uniform. Say $M_2=U_{r-1,n}$. Then $B(M_1)$ is the set of all $r$ element sets of an $n$ element set and so $M_1=U_{r,n}$, proving the claim.