The other answers and comments provided examples where all but one of the minors vanish, so you see that you need to check all of them. Let me put this in different context. Define the $m \times n$ generic matrix $X$ over a field $k$ (or a commutative ring) to have its entries $x_{i,j}$, where the $x_{i,j}$ are variables in a polynomial ring $k[x_{i,j}]$. The determinantal variety of this matrix is the ideal generated by the $r \times r$ minors ($r$ being the size you're interested in).
Your question amounts to asking if these $\binom{m}{r} \binom{n}{r}$ generators are linearly independent (over $k$). Let me explain why they cannot be using some representation theory. There is an action of $G = {\bf GL}_n(k) \times {\bf GL}_m(k)$ on the generic matrix via $(A,B) \cdot X = AXB^{-1}$ which preserves rank. Hence the set of conditions for this matrix to have rank $< r$ must be preserved by $G$. We can rewrite $k[x\_{i,j}]$ as $k[V^* \otimes W] \cong {\rm Sym}(V \otimes W^\*)$ where $V$ and $W$ are $k$-vector spaces of ranks $m$ and $n$, respectively. The $r \times r$ minors are polynomials of degree $r$, hence sit inside of ${\rm Sym}^r(V \otimes W^\*)$. In fact, they must span a $G$-submodule. Furthermore, this $G$-submodule is $\bigwedge^r V \otimes \bigwedge^r W^\*$ (which can be seen in various ways), which we know has dimension $\binom{m}{r} \binom{n}{r}$. Hence they are linearly independent (as polynomials over $k$). This explains why you will always find examples where all but 1 of the minors can vanish (when we specialize the $x_{i,j}$ to specific values in $k$).
Another question you can ask is whether this ideal is radical. The answer is yes, but this requires some more work. So there are no shortcuts! Well, you can do Gauss-Jordan elimination. That would be faster.