Proving that there is no matrix $B$ for such $By = x$

There are many possible matrices $B$ statisfying $By=x$. Note that $B$ represents a linear transformation that maps $y$ to $x$. To define such linear map and computing its matrix, extend $\{y\}$ to a basis $\{y, v_2,v_3,v_4\}$ of $\mathbb{R}^4$, which is possible since $y\neq 0$. Then define a map $f:\mathbb{R}^4\to\mathbb{R}^6$ by setting $f(y)=x$ and then decide where you want to send $f(v_i)$ (for instance $f(v_i)=0$, to make the matrix very easy to compute). After this, you just have to compute the matrix $B$ of $f$ and by definition it will satisfy $By=x$.

The matrix $B$ need not be of the form $(A^TA)^{-1}A^T$, that is where your argument fails.

Edit: I see that in your post you say "associate with $A$", I undertand that in this context (real vector spaces) this just mean $A^T$. In that case you are right, $B$ is not of the form $A^T$, but in this case since you know what $A$ is like, you can just check that $A^Ty\neq x$.

An extra comment on your justification is that $Mx=Ly$ does not imply $x=M^{-1}Ly$, as the inverse may not exist, as it is in your case. For example, take the vectors $x=(1,1)$ and $y=(1,0)\in\mathbb{R}^2$. If you take $M$ and $L$ to be both the matrix associated to the projection $p_1(a,b)=a$, you will get $Mx=Ly$, but $M^{-1}$ does not exist because the map is not injective. In particular this means that even if you were right, your proof was not.