I posted this question on a different site a couple of years ago.  Eventually I found that a book of T.Y. Lam has a very nice treatment.  Here is the writeup I posted on the other site:

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After paging through several algebra books, I found that T.Y. Lam's GTM "Lectures on 
Rings and Modules" has a beautiful treatment of this question.

The above property of a (possibly noncommutative) ring is called the "strong rank 
condition." It is indeed stronger than the corresponding statement for 
surjections ("the rank condition") which is stronger than the isomorphism version 
"Invariant basis number property". However, in fact it _is_ the case that all 
commutative rings satisfy the strong rank condition. Lam gives two proofs 
[pp. 12-16], and I will now sketch both of them.

First proof: 

Step 1: The result holds for (left-) Noetherian rings. For this we establish:

Lemma: Let M and N be (left-) A-modules, with N nonzero. If the direct sum 
M + N can be embedded in M, then M is not a Noetherian A-module.

Proof: (I will use + to denote direct sum) By hypothesis M has a submodule 
M_1 + N_1, with M_1 isomorphic to M and N_1 isomorphic to N. But we can also 
embed M+N in M_1, meaning that M_1 contains a submodule M_2 + M_2 with M_2 isomorphic 
to M and N_2 isomorphic to N. Continuing in this way we construct an ascending 
chain of submodules N_1, N_1 + N_2,..., contradiction.

So if A is (left-) Noetherian, apply the Lemma with M = A^n and N = A^{m-n}. 
M is a Noetherian A-module, and we conclude that A^m cannot be embedded in A^n.

Step 2: We do the case of a commutative, but not necessarily Noetherian, ring. 
First observe that, defining linear independent subsets in the usual way, the 
strong rank condition precisely asserts that any set of more than n elements in A^n 
is linearly dependent. Thus a ring A satisfies the strong rank condition iff: for 
all m > n, any homogeneous linear system of n linear equations and m unknowns has a 
nonzero solution in A.

So, let MX = 0 be any homogeneous linear system with coefficient matrix M = (m_{ij})
1 <= i <= n, 1 <= j <= m. We want to show that it has a nonzero solution in A. But 
the subring A' = Z[a_{ij}], being a quotient of a polynomial ring in finitely many 
variables over a Noetherian ring, is Noetherian (by the Hilbert basis theorem), so 
by Step 1 there is (even) a nonzero solution (x_1,...,x_m) in (A')^m.

This makes one wonder if it is necessary to consider the Noetherian case separately, 
and it is not. Lam's second proof comes, he says, from Bourbaki's _Algebra_ 
(unfortunately he does not give a precise reference). It uses the following 
elegant characterization of linear independence in free modules:

Theorem: A subset {u_1,...,u_m} in M = A^n is linearly independent iff 

If a in A is such that a*(u_1 ^ ... ^ u_m) = 0, then a = 0. 

Here u_1 ^ ... ^ u_m is an element of the exterior power Lambda^m(M). 

(I will omit the proof here; the relevant passage is reproduced on Google books.)

This gives the result right away: if m > n, Lambda^m(A^n) = 0.