= Solution
Write $R=\mathbb Z[\sqrt{-13}]$ and $I=(2,1+\sqrt{-13})$. The map
$$
R\longrightarrow\mathbb F_2,
\qquad a+b\sqrt{-13}\longmapsto a+b\pmod2
$$
is 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, then
$$
a+b\sqrt{-13}=b(1+\sqrt{-13})+(a-b),
$$
and $a-b$ 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 index
$$
|N(\alpha)|=|\alpha\overline\alpha|=a^2+13b^2.
$$
Thus principality would require $a^2+13b^2=2$. This <Diophantine equation> has no integer solution: $b=0$ would give $a^2=2$, while $b\ne0$ makes the left side at least $13$. Therefore
$$
\boxed{(2,1+\sqrt{-13})\text{ is not principal}}.
$$
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