Here is a general proof for all parameters in an open domain.
$$dr = adt+bdW:=r\big(k(\theta-x)+\frac12\sigma^2\big)dt+\sigma rdW.$$
Let
$$u(r(s),s):=e^{-\int_t^sr}B(r(s),s,T)=:\phi(s) B.$$
Then
$$u(r(t),t)=\mathbf E\big[u(r(s),s)\big|r(t)\big],\, \forall t<s. \tag{1}$$
So, by Ito's Lemma,
\begin{align}
du(r(s),s) &= Bd\phi +\phi dB \\
&= \phi \bigg(-rB+\frac{\partial B}{\partial s}ds+\frac{\partial B}{\partial r}dr+\frac12\frac{\partial^2 B}{\partial r^2}(dr)^2\bigg) \\
&= \phi \bigg[\bigg(-rB+\frac{\partial B}{\partial s}+\frac{\partial B}{\partial r}a+\frac12\frac{\partial^2 B}{\partial r^2}b^2\bigg)ds+\frac{\partial B}{\partial r}bdW\bigg] \\
&=: \phi\,(fds+gdW_s).
\end{align}
We see from Eq. (1) $\mathbf E\big[u(r(s),s)\big|r(t)\big]$ is constant with respect to $s$. So
$$0=\frac{d\mathbf E\big[u(r(s),s)\big|r(t)\big]}{ds}\bigg|_{s=t}=f(r(t),t) \tag{2}$$
by Equation (1).
Suppose $B$ is affine. Substitute into $\frac{f}{B}$ the affine expression for $B(r,t,T)$ and the expression of $a$ and $b$, we have by Equation (2)
$$A'-\Big(C'+\Big(k\theta+\frac{\sigma^2}2\Big)C-1\Big)e^{X_t}+kCX_te^{X_t}+\frac{(\sigma C)^2}{2}e^{2X_t}=0,\quad\forall X_t\in R,$$
where $'$ denotes partial derivative with respect to $t$ (denoting the first variable). By taking derivatives with respect to $X_t$ or Taylor expanding $e^{X_t}$, we see $(1,e^{X_t},X_te^{X_t},e^{2X_t})$ is linearly independent. So all factors in front of those terms vanish. This is possible only when $k=\sigma=0,\,C(t,s)=s-t$ and $A(t,s)=0$.
