 Solution

Suppose for a contradiction that $m>n$. Compose the given injective module homomorphism with the standard injection $B^n\hookrightarrow B^m$ that appends $m-n$ zero coordinates. This gives an injective endomorphism $T$ of the finite free module $B^m$ whose matrix has a zero final row.

Its characteristic polynomial has zero constant term, so the Cayley-Hamilton theorem givesInjectivity lets us cancel $T$. Repeating this argument eventually gives the identity endomorphism equal to zero. That would imply $B^m=0$, contrary to $B\ne 0$. Hence $m\le n$. This proves the rank inequality for an injection of finite free modules.