First,
$$
X_t=\rho X_{t-1}+\epsilon_t=\rho^2 X_{t-2}+\rho\epsilon_{t-1}+\epsilon_t=\ldots=\sum_{s=0}^{t-1}\rho^s\epsilon_{t-s},
$$
and
$$
\begin{bmatrix}
X_1 \\
X_2 \\
\vdots \\
X_t
\end{bmatrix}=
\begin{bmatrix}
1 & 0 & \cdots & 0 \\
\rho & 1 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
\rho^{t-1} & \rho^{t-2} & \cdots &1
\end{bmatrix}
\begin{bmatrix}
\epsilon_1 \\
\epsilon_2 \\
\vdots \\
\epsilon_t
\end{bmatrix}.
$$
Therefore, $[X_1,\cdots, X_t]^{\top}\sim N(0, \sigma^2 R_{\rho}R_{\rho}^{\top})$, where $R_{\rho}$ is the $t\times t$ matrix in the preceding equation.
Let $\vec{x}\equiv [x_1,\ldots,x_t]^{\top}$. Then the likelihood function for $(\sigma,\rho)$ is given by
$$
\mathcal{L}(\sigma,\rho;\vec{x})=(2\pi\sigma^2)^{-k/2}e^{-\frac{1}{2\sigma^2}\vec{x}^{\top}\left(R_{\rho}R_{\rho}^{\top}\right)^{-1}\vec{x}}
$$
because $\operatorname{det}(R_{\rho}R_{\rho}^{\top})=1$. The quadratic form appearing in $\exp(\,\cdot\,)$ can be written as
$$
\vec{x}^{\top}\left(R_{\rho}R_{\rho}^{\top}\right)^{-1}\vec{x}=(1+\rho^2)\sum_{s=1}^{t-1}x_s^2+x_t^2-2\rho\sum_{s=1}^{t-1}x_sx_{s+1}.
$$
Now use the Neyman-Pearson factorisation criterion.