So obviously $6$ isn't a quadratic residue $mod(7)$ thus there are no zeros in $\Bbb Z_7$. So what I did next is considered considered the field $\Bbb Z_7[x]/<x^2+1>$ obviously $x+<x^2+1>$ is a zero for the polynomial so we will denote that by $i$, keep in mind that $i^2=6$. My question is really about the field $\Bbb Z_7(i)$, that is the smallest field containing all of $\Bbb Z_7$ and $i$. Well since $i^2=6$ this field has all its elements of the form $a+bi$ where $a,b \in \Bbb Z_7$ so that field contains only 14 elements which isn't a power of a prime. I must be doing something wrong since I am a bit rusty on algebra, can someone please point out my mistake.
$(a+bi)^2 = a^2 - b^2 + 2abi = -1$ only iff $2ab = 0$, which gives us that either $a = 0$ or $b=0$. You have excluded the options where $b=0$, which leaves us with $a = 0$. Here we get the solutions $b = 6$ and $b = 1$.