For this question, I simply need some clarification. I understand that in $Z$, the equation $x^2 = a$ has a solution only when $a$ is a perfect square (and hence positive) or when $a = 0$.
However, I simply am not able to understand how this works in $\Bbb Z/7\Bbb Z$.
Just compute the squares in ${\Bbb Z}_7$: $1^2=1$, $2^2=4$, $3^2= 2$, $4^2=2$, $5^2=4$, $6^2=1$. So for $a=1,2,4$ you have solutions $x$.
An integer $a$ is a quadratic residue modulo $n$ if it is congruent to a perfect square modulo $n$; i.e., if there is an integer $x$ such that $$x^2\equiv a\mod n.$$ Otherwise, $a$ is a quadratic nonresidue modulo $n$.
There is a whole theory behind this in number theory. Please have a look at https://en.wikipedia.org/wiki/Quadratic_residue