Hint
$\mathbb Z[i]$ is a PID. If $(7)$ or $(13)$ is not maximal, it can be included in some larger ideal $(m)$.
This means that $m|7$ or $m|11$ in $\mathbb Z[i]$. Now all you need is to figure out if you can find $a+bi$ and $c+di$ which are not units and
$$(a+bi)(c+di)=7$$
respectively
$$(a+bi)(c+di)=13$$
Your ring $\mathbb{Z}_n[i]$ can be seen as the quotient $$\Bbb Z_n [i] \cong \Bbb Z[i] / (n).$$ One of the isomorphism theorems, which can be found on every algebra book, says that the ideals of a quotient $R/A$ are in bijection with the ideals of $R$ that contain $A$, and the bijection is given by $$B \mapsto B/A = \{b+A : b \in B\}.$$Because $\Bbb Z[i]$ is a PID, every ideal that contains $(n)$ is of the form $(m)$, where $m \mid n$. If $$n=p_1^{k_1} \ldots p_r^{k_r}$$is the decomposition of $n$ as product of primes (in $\Bbb Z [i]$), then any ideal containing $(n)$ is of the form $(p_1^{s_1} \ldots p_r^{s_r})$, where $s_j \leq k_j$ and, in particular, there are $(k_1+1) \ldots (k_r+1)$ different ideals, which, in turn, correspond to different ideals of $\mathbb{Z}_n[i]$.\
Note that there is nothing particular of $\Bbb Z [i]$ apart from being a PID that we are using here. The same method proves how to find all the ideals in the quotient of a PID by one of its ideals, as you would do with $\Bbb Z _n$, but now taking into account that you need to factorise in $\Bbb Z[i]$ so, for example, $5$ is not prime because $5=(2+i)(2-i)$.
As it has been said in the comments, every ideal of $\mathbb{Z}_n[i]$ is principal. This is a general result. If you look back to the correspondence of ideals in a quotient, if $B$ is generated by $b$, then $B/A$ is generated by $b+A$.
Best Answer
$\mathbb Z[i]$ is a Euclidean domain, and thus a principal ideal domain. So every ideal is generated by a single element, this should help you give a description of the proper ideals. The prime ideals are related then to the prime elements which are known as Gaussian primes. Have a look at this for more information.