Well, to me your attempt is as intuitive as it gets. Just a remark: you assumed that $m\neq n$ to conclude that $x$ can't be in the column space or left null space, which a priori is not given (at least, it wasn't stated in the question). But this is ok, since, if $m=n$ and $A$ has independent columns, $A$ would be an invertible square matrix, and $A^TAx=A^Tb\implies Ax=b$, but, since $b$ is assumed not to be in the column space of $A$, this case doesn't happen.
Maybe what you're asking for is a more explicit explanation? If so, Let $A=\begin{bmatrix}L_1 \\ \vdots\\ L_m\end{bmatrix}$, where $L_i$ is the $i$-th line, a $1\times n$ vector. If $b=\begin{bmatrix} b_1 \\ \vdots\\ b_m\end{bmatrix}$, we have that
\begin{align*} A^TAx=A^Tb&\implies \begin{bmatrix} L_1 & \cdots & L_m\end{bmatrix}\begin{bmatrix}L_1 \\ \vdots\\ L_m\end{bmatrix}\textbf{x}=\begin{bmatrix} L_1 & \cdots & L_m\end{bmatrix}\begin{bmatrix} b_1 \\ \vdots\\ b_m\end{bmatrix}\\ &\implies(L_1\cdot L_1+\dots+L_n\cdot L_n)\textbf{x}=b_1L_1+\dots+b_mL_m\\ &\implies\textbf{x}=\dfrac{1}{L_1\cdot L_1+\dots+L_n\cdot L_n}(b_1L_1+\dots+b_mL_m) \end{align*}
So $\textbf{x}$ is indeed in the row space of $A$ (notice that the denominator of the fraction is clearly non-zero).
