Show that the ideal $I = (2,1+\sqrt{-13})$ is a maximal ideal in $\mathbb{Z}[\sqrt{-13}]$? Is it principal? 
Show that the ideal $I = (2,1+\sqrt{-13})$ is a maximal ideal in $\mathbb{Z}[\sqrt{-13}]$? Is it principal?

To show that the ideal is maximal, I know that $I \text{ maximal} \iff \mathbb{Z}[\sqrt{-13}]/I \text{ is a field}\iff \text{only ideals of } \mathbb{Z}[\sqrt{-13}]/I \text{ are } 0,\mathbb{Z}[\sqrt{-13}]/I$.
I have determined that $I = \{(2a+c-13d) + (2b+c+d)\sqrt{-13} : a,b,c,d \in\mathbb{Z}\}$.
I now do not know how to continue showing that I is maximal.
To show $I$ is not principal, lets assume $I = (a)$ for some ring element $a$.
Then since $2\in(a)$, we have $N(a) | 4$ so $N(a) = 1,2,4$.
But $x^{2}+13y^{2} =1$ means $a=1\implies I = \mathbb{Z[\sqrt{-13}]}$ which is a contradiction since $1\not\in I$.
$N(a) = 2$ has no solutions. So $N(a) = 4 \implies N(a)|13$ which is another contradiction (since $\sqrt{-13} \in I$)
 A: Use your first equivalence: $I$ is maximal if and only if ${\mathbb Z}[\sqrt{-13}]/I$ is a field. The trick now is to rewrite the quotient ring until you see that it's a field:
\begin{equation*}
{\mathbb Z}[\sqrt{-13}]/(2, 1 + \sqrt{-13})
\cong
{\mathbb Z}[x]/(2, 1 + x, x^2 + 13)
\cong
{\mathbb F}_2[x]/(1 + x, x^2  + 1)
\cong
{\mathbb F}_2.
\end{equation*}
Each of the isomorphisms requires justification, but every time it's the 'obvious' map that gives the isomorphism and it's the same type of reasoning for each of this type of problems.
A: Let $I\subset J$ be a larger ideal and $a+b\sqrt{-13}\in J-I$ since $2\in I$ we may assume that $a,b =\pm 1$. Now $\pm(1\pm\sqrt{-13})\in I$, so this leaves the possibilities $\pm 1$ and $\pm \sqrt{-13}$ both of which entail that $1\in J$ and so $J$ is not proper. 
For non principal, if $2=xy$ then as you observe $4=Nx Ny$ and so $Nx=1$, or $Ny=1$ or $Nx=Ny=2$. The latter is impossible and the former implies that one of $x$ or $y$ is in $\mathbb{Z}$, also impossible. 
