 Solution

Write $R=\mathbb{Z}[\sqrt{-13}]$ and $I=(2,1+\sqrt{-13})$. The mapis a surjective ring homomorphism: in ${\mathbb{F}}_2$, the image of $\sqrt{-13}$ is $1$ and $1^2=-13=1$. Its kernel is exactly $I$. Indeed, both generators lie in the kernel; conversely, if $a+b$ is even, thenand $a-b$ is even. Hence $R/I\cong {\mathbb{F}}_2$ and the index of a subgroup is $[R:I]=2$.

If $I=(\alpha )$ were a principal ideal, multiplication by $\alpha =a+b\sqrt{-13}$ would identify its additive lattice with a sublattice of indexThus principality would require $a^2+13b^2=2$. This Diophantine equation has no integer solution: $b=0$ would give $a^2=2$, while $b\ne 0$ makes the left side at least $13$. Therefore