Let $K$ be the field you want to construct and let $\Bbb Q(i)$ the field of complex numbers of the form $r+si$ with both $r$ and $s$ in $\Bbb Q$.
First thing you need to convince yourseld that $K\subset{\Bbb Q}(i)$. If you take a fraction $\frac{a+bi}{c+di}$ with $a,b,c,d\in\Bbb Z$ you can rewrite it as
$$
\frac{a+bi}{c+di}=\frac{ac+bd}{c^2+d^2}+\frac{bc-ad}{c^2+d^2}i
$$
(why?) and you're done.
On the other hand, if $\frac mn+\frac pqi\in\Bbb Q(i)$, we can rewrite this as
$$
\frac{mq+npi}{nq}
$$
where both numerator and denominator are Gaussian integers. Thus you also have $\Bbb Q(i)\subset K$ and you're finished.
Yes, if $n$ is a rational integer, then $\mathbb{Z}[i]/n\mathbb{Z}[i]$ is a field if and only if it is an integral domain, if and only if $n$ is a prime number which is $\equiv 3 \pmod{4}$. You can argue this using some elementary number theory and some results about unique factorization domains (as well as the fact that $\mathbb{Z}[i]$ is a unique factorization domain). I'll show that if $n$ is not a prime number which is $3 \pmod{4}$, then that ring is not an integral domain.
Writing $n$ as a product of prime numbers and using the Chinese remainder theorem, we can reduce to the case where $n$ is a prime power, say $n = p^m$ for some $m \geq 1$.
Obviously, $\mathbb{Z}[i]/ p^m\mathbb{Z}[i]$ is not an integral domain when $m \geq 2$, since the image of $p$ there is nonzero and nilpotent. So $m$ must be $1$.
If $p = 2$, then $\mathbb{Z}[i]/2\mathbb{Z}[i]$ is not an integral domain, because the images of $1+i$ and $1-i$ are nonzero, yet their product is.
If $p \equiv 1 \pmod{4}$, then by a result from elementary number theory, $p$ can be written as a sum of two squares: $p =a^2 + b^2$. Actually, you can argue this directly from the fact that $\mathbb{Z}[i]$ is a UFD. Then $p = (a+bi)(a-bi)$, and again the images of $a+bi$ and $a-bi$ in the quotient ring are nonzero, but their product is.
Best Answer
If $z,w\in\mathbb{Z}[i]$ are such that $zw=1$ (i.e. $z$ is a unit and $w$ its inverse), then $|z|^2|w|^2=|zw|^2=1$, or
$$(a^2+b^2)(c^2+d^2)=1, \quad z=a+bi,\; w=c+di.$$
Now $a,b,c,d$ are all integers, so $a^2+b^2$ and $c^2+d^2$ must both be nonnegative integers, which must both equal exactly $1$ and no greater in order to multiply to $1$ in the integers. And if $a^2+b^2=1$, we have $a^2$ and $b^2\le1$. Check by hand the only solutions here correspond to $(a,b)=(\pm1,0)$ or $(0,\pm1)$.