$N$ is flat if and only if when $f: M' \rightarrow M$ is injective then $f \otimes \text{id}_N : M' \otimes N \rightarrow M \otimes N$ is injective ($N,M,M'$ are all $A$-modules). So consider $f: M' \rightarrow M$ injective and tensor by the $A$-module $A[x]$. We get $f \otimes \text{id}_{A[x]} : M' \otimes A[x] \rightarrow M \otimes A[x]$. To check if this is injective, we consider its kernel: $$ \begin{align} \ker(f \otimes \text{id}_{A[x]}) &= \ker(f) \otimes A[x] + M' \otimes \ker(\text{id}_{A[x]}) \\ &= 0 \otimes A[x] + M' \otimes 0 \\ &=0+0 \\ &=0. \end{align} $$ Here I have used the rule for kernels of tensor products of maps, that $f$ is injective and that the identity map is injective. So the tensored map is injective and so $A[x]$ is flat.
I'm not sure about this because it seems that you could prove any module is flat this way, and obviously not all modules are flat. Would anyone be able to shed some light on where my reasoning has gone wrong? I have indeed seen this way to do this question, but I'd still like to know why this method is wrong. Many thanks.
The formula for the kernel of the tensor of maps that you refer to is only valid when both maps are surjective. Here's a reference (Theorem 2.19) to a proof in some free notes of Keith Conrad, and notably surjectivity plays a big role in the reasoning by allowing us to describe the form of elementary tensors in the image.
As you mentioned, if the kernel identity applied across the board, or even just to injective maps, or even just to injective maps tensored with isomorphisms, then every module would be flat, which is silly. Indeed the tensor of injective maps need not be injective, and modules need not be flat.
In fact, every integral domain which is not a field has lots of non-flat modules over it which are very easy to construct.
For a classic example with the integers, take $\iota: \mathbb{Z} \rightarrow \mathbb{Q}$ to be inclusion and $\text{id}: \mathbb{Z}/n\mathbb{Z} \rightarrow \mathbb{Z}/n\mathbb{Z}$, considering all as $\mathbb{Z}$-modules. Sure enough, $(\iota \otimes \text{id})(1 \otimes \bar{1})$. Both maps are injective and $1 \otimes \bar{1}$ is nonzero in $\mathbb{Z} \otimes \mathbb{Z}/n\mathbb{Z}$ but the tensor of maps is not injective since $(\iota \otimes \text{id})(1 \otimes \bar{1}) = \frac{1}{n} \otimes \bar{0}$ in $\mathbb{Q} \otimes \mathbb{Z}/n\mathbb{Z}$.