Just to recall some basic stuff: Suppose $Y_i = \alpha + \beta x_i + \varepsilon_i$ for $i=1,\dots,n$ and $\varepsilon_i \sim N(0,\sigma^2)$ and these are independent. We will observe the $x$'s and $y$'s but not $\alpha$, $\beta$, or $\varepsilon_i$. The $Y$'s are random only because the $\varepsilon$'s are random. These are weaker assumptions than that we have $(x,y)$ pairs that are *jointly* normally distributed.

Let
$$
X = \begin{bmatrix}
1 & x_1 \\ \vdots & \vdots \\ 1 & x_n
\end{bmatrix}
$$
be the "design matrix" (so called because if the experimenter can choose the $x$ values, then this is how the experiment is designed).

Then the least-squares estimates of $\alpha$ and $\beta$ are given by
$$
\begin{bmatrix} \hat\alpha \\ \hat\beta \end{bmatrix} = (X^T X)^{-1}X^T Y
$$
and therefore the probability distribution of the least-squares estimators is given by
$$
\begin{bmatrix} \hat\alpha \\ \hat\beta \end{bmatrix} \sim N\left( \begin{bmatrix} \alpha \\ \beta \end{bmatrix}, \sigma^2 (X^T X)^{-1} \right)
$$
(the variance is thus of course a $2\times 2$ positive-definite matrix). The predicted $y$-value for a given $x$ value is therefore $\hat\alpha + \hat\beta x$, and this therefore has a probability distribution given by
$$
\hat y = \hat\alpha + \hat\beta x = \begin{bmatrix}1, & x\end{bmatrix} \begin{bmatrix} \hat\alpha \\ \hat\beta \end{bmatrix} \sim N\left( \begin{bmatrix}1, & x\end{bmatrix} \begin{bmatrix} \alpha \\ \beta \end{bmatrix}, \sigma^2 \begin{bmatrix} 1, & x \end{bmatrix} (X^T X)^{-1} \begin{bmatrix} 1 \\ x \end{bmatrix} \right)
$$
$$
= N\left( \alpha + \beta x, \frac{\sigma^2}{n}\frac{\sum_i (x_i - x)^2}{\sum_i (x_i - \overline{x})^2} \right).
$$
(OK, check my algebra here; it's trivial but laborious.)

There's a simple geometric intuition behind the dependence of the variance on $x$ and in particular the fact that the variance is smallest when $x=\overline{x}$, so think about that too.

Now $\sigma^2$ must be estimated based on the data. The *errors* $y_i - (\alpha + \beta x_i)$ are unobservable but but the *residuals* $y_i - (\hat\alpha + \hat\beta x_i)$ (i.e. the *estimated* errors) are the components of the random vector
$$
\hat\varepsilon = (I - H)Y = (I - X(X^T X)^{-1} X^T)Y.
$$
("H" stands for "hat", for reasons that should be apparent.) It is easy to see that the $n\times n$ hat matirx $H = X(X^T X)^{-1} X^T$ is the matrix of the orthogonal projection of rank $2$ onto the 2-dimensional column space of $X$. And $I - H$ is the rank-$(n-2)$ projection onto the orthogonal complement of that space. Diagonalized, this matrix just has $n-2$ instances of 1 on the diagonal and 0 in the other two positions. Therefore
$$
\frac{\hat\sigma^2}{\sigma^2} = \frac{\| \hat\varepsilon \|^2}{\sigma^2}
$$
is distributed like a sum of squares of $n - 2$ independent $N(0,1)$ random variables. It therefore has a chi-square distribution with $n-2$ degrees of freedom.

Finally, we need this: $\hat\varepsilon$ and $\begin{bmatrix}\hat\alpha \\ \hat\beta \end{bmatrix}$ are probabilistically independent. This is true because both are linear transformations of the same vector of independent identically distributed normal random variables and their covariance vanishes:
$$
\operatorname{cov}\left(\hat\varepsilon, \begin{bmatrix}\hat\alpha \\ \hat\beta \end{bmatrix} \right) = \operatorname{cov}\left( (I - H)Y , (X^T X)^{-1}X^ T \right)
$$
$$
= (I - H) \operatorname{cov}(Y,Y) X(X^T X)^{-1} = \sigma^2 (I - H) X(X^T X)^{-1}$
$$
and this is the $n\times 2$ zero matrix, by definition of $H$.

Now all our lemmas are in place and we can draw some conclusions:

Firstly
$$
\frac{\hat y - (\alpha + \beta x)}{\sqrt{ \frac{\sigma^2}{n}\frac{\sum_i (x_i - x)^2}{\sum_i (x_i - \overline{x})^2} }} \sim N(0,1).
$$

Hence if $\sigma$ were miraculously known, we could say that
$$
\hat y \pm A \sqrt{ \frac{\sigma^2}{n}\frac{\sum_i (x_i - x)^2}{\sum_i (x_i - \overline{x})^2} }
$$
are the endpoints of a 90% confidence interval for $\alpha + \beta x$ if $\pm A$ are the endpoints of the interval above which lies 90% of the area under the bell-curve.

But $\sigma$ is not known. Since $\hat\sigma^2$ is indpendent of the random variable in the numerator and has a chi-square distribution with $n-2$ degrees of freedom, we can put $\hat\sigma$ in place of $\sigma$ and instead of the normal distribution use the Student's t-distribution with $n-2$ degrees of freedom.

That's the conventional frequentist confidence interval.

For the prediction interval, just remember that the new value of $Y$ is independent of those we used above, so the variance of the difference between that and the predicted value is $\sigma^2$ plus the variance of the predicted value, found above.