= Solution
Let $\operatorname{RSS}=\sum_i(Y_i-\widehat\alpha-\widehat\beta x_i)^2$. Differentiating the <log-likelihood> in $v=\sigma^2$ gives
$$
\frac{\partial\ell}{\partial v}=-\frac n{2v}+\frac{\operatorname{RSS}}{2v^2}.
$$
For $\operatorname{RSS}>0$ this changes from positive to negative at $v=\operatorname{RSS}/n$, so
$$
\boxed{\widehat\sigma^2=\frac{\operatorname{RSS}}n.}
$$
To find its bias, substitute the model and use the orthogonal fit directions $\mathbf1$ and $\mathbf x$, obtaining
$$
\operatorname{RSS}=\sum_i\varepsilon_i^2-n\bar\varepsilon^2-\frac{(\sum_i x_i\varepsilon_i)^2}{S_{xx}}.
$$
The three expectations are $n\sigma^2$, $\sigma^2$, and $\sigma^2$, respectively: $\operatorname{Var}(\bar\varepsilon)=\sigma^2/n$ and $\operatorname{Var}(\sum_i x_i\varepsilon_i)=\sigma^2S_{xx}$. Hence $\mathbb E\operatorname{RSS}=(n-2)\sigma^2$, giving the <unbiased estimator>
$$
\boxed{s^2=\frac{\operatorname{RSS}}{n-2}=\frac n{n-2}\widehat\sigma^2.}
$$
With $n>2$, a zero <residual sum of squares> is a probability-zero event in the model. For that exceptional dataset the likelihood is unbounded as $\sigma^2\downarrow0$, rather than having a maximum at a positive <variance>. In the intercept-only case $S_{xx}=0$, the corresponding unbiased multiplier is $n/(n-1)$ for $n>1$.
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