I found the answer in this book (in Section $6.4.14$, “Determinants, Ranks and Linear Equations”). I'd tried using a similar Laplace expansion myself but was missing the idea of using the largest dimension at which the minors are not all annihilated by the same non-zero element. I'll try to summarize the argument in somewhat less formal terms, omitting the tangential material included in the book.
Let $A$ be an $m\times n$ matrix over a commutative ring $R$. We want to find a condition for the system of equations $Ax=0$ with $x\in R^n$ to have a non-trivial solution. If $R$ is a field, various definitions of the rank of $A$ coincide, including the column rank (the dimension of the column space), the row rank (the dimension of the row space) and the determinantal rank (the dimension of the lowest non-zero minor). This is not the case for a general commutative ring. It turns out that for our present purposes a useful generalization of rank is the largest integer $k$ such that there is no non-zero element of $R$ that annihilates all minors of dimension $k$, with $k=0$ if there is no such integer.
We want to show that $Ax=0$ has a non-trivial solution if and only if $k\lt n$.
If $k=0$, there is a non-zero element $r\in R$ which annihilates all matrix elements (the minors of dimension $1$), so there is a non-trivial solution
$$A\pmatrix{r\\\vdots\\r}=0\;.$$
Now assume $0\lt k\lt n$. If $m\lt n$, we can add rows of zeros to $A$ without changing $k$ or the solution set, so we can assume $k\lt n\le m$. There is some non-zero element $r\in R$ that annihilates all minors of dimension $k+1$, and there is a minor of dimension $k$ that isn't annihilated by $r$. Without loss of generality, assume that this is the minor of the first $k$ rows and columns. Now consider the matrix formed of the first $k+1$ rows and columns of $A$, and form a solution $x$ from the $(k+1)$-th column of its adjugate by multiplying it by $r$ and padding it with zeros. By construction, the first $k$ entries of $Ax$ are determinants of a matrix with two equal rows, and thus vanish; the remaining entries are each $r$ times a minor of dimension $k+1$, and thus also vanish. But the $(k+1)$-th entry of this solution is non-zero, being $r$ times the minor of the first $k$ rows and columns, which isn't annihilated by $r$. Thus we have constructed a non-trivial solution.
In summary, if $k\lt n$, there is a non-trivial solution to $Ax=0$.
Now assume conversely that there is such a solution $x$. If $n\gt m$, there are no minors of dimension $n$, so $k\lt n$. Thus we can assume $n\le m$. The minors of dimension $n$ are the determinants of matrices $B$ formed by choosing any $n$ rows of $A$. Since each row of $A$ times $x$ is $0$, we have $Bx=0$, and then multiplying by the adjugate of $B$ yields $\det B x=0$. Since there is at least one non-zero entry in the non-trivial solution $x$, there is at least one non-zero element of $R$ that annihilates all minors of size $n$, and thus $k\lt n$.
Specializing to the case $m=n$ of square matrices, we can conclude:
A system of linear equations $Ax=0$ with a square $n\times n$ matrix
$A$ over a commutative ring $R$ has a non-trivial solution if and only
if its determinant (its only minor of dimension $n$) is annihilated by
some non-zero element of $R$, that is, if its determinant is a zero divisor or zero.
As commented by Jyrki Lahtonen, the statement is true and it is immediately implied by the following relation between the minors of $A^{-1}$ and the minors of $A$.
Proposition: If $A$ is an invertible $n\times n$ matrix, and if $i_1,\dots,i_n$ and $j_1,\dots,j_n$ be two permutations of $1,\dots,n$, then the minor of $A$ corresponding to rows $i_1,\dots,i_k$ and columns $j_1,\dots,j_k$, denoted by $d$, and the minor of $A^{-1}$ corresponding to rows $j_{k+1},\dots,j_n$ and columns $i_{k+1},\dots,i_n$, denoted by $d'$,satisfy that
$$d=\pm d'\det A.$$
Proof: Let $e_1,\dots,e_n$ be a basis of $\mathbb{R}^n$, and let $f_i= A e_i$, $i=1,\dots,n$. Then on the one hand,
$$\omega:=Ae_{j_1}\wedge\cdots\wedge Ae_{j_k}\wedge e_{i_{k+1}}\wedge \cdots\wedge e_{i_n}=\pm d\cdot e_1\wedge\cdots\wedge e_n,$$
on the other hand,
$$\omega=f_{j_1}\wedge\cdots\wedge f_{j_k}\wedge A^{-1}f_{i_{k+1}}\wedge \cdots\wedge A^{-1}f_{i_n}=\pm d'\cdot f_1\wedge\cdots\wedge f_n.$$
Since
$$f_1\wedge\cdots\wedge f_n=\det A\cdot e_1\wedge\cdots\wedge e_n,$$
the conclusion follows.
Best Answer
Well spotted. In a case like this, it's a good idea to check the article's history (using the "View history" link at the top). In the present case, the error was introduced only two days ago by an anonymous user in this edit (which I just reverted).