Just do it.
All integers are either congruent to $0, \pm 1, \pm 2, \pm 3 \mod 7$. So any integer square with be congruent to $0, 1 , 4, 9\mod 7$ or in other words to $0,1,4,2 \mod 7$.
That's all there is to it.
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Big picture: For any positive integer $n$, we know by division algorithm that for any integer $m$ there are unique integers $q$ and $r$ so that $m = q*n + r; 0\le r < n$. In other words $r$ is the remainder of dividing $m$ by $n$.
This means each integer will have a unique remainder $r$ and as there are exactly $n$ possible values for $r$, we can classify all the integers into $n$ different class types depending upon which of $n$ possible remainders it has. For example: in modulo $7$ we can separate the integers into $7$ classes. One class $\{...-11,-4, 3, 10, 17,....\}$ are all the integers that have $3$ as a remainder. $\{....,-10,-3, 4, 11, 18,...\}$ are all the integers that have $4$ as a remainder.
$a \equiv b \mod n$ means that $a$ and $b$ both belong to the same class. $a = qn + r$ and $b=pn + r$ for possibly different $p,q$ but both the same $r$. Now keep in mind, ALL integers belong to one of only $n$ possible classes.
Big result: Belonging to modulo classes is closed/maintained under addition, multiplication and raising to powers. That is: if $a \equiv b \mod n$ and $c \equiv d \mod n$. Then $a+c \equiv b+d \mod n$.
Proof: If $a$ and $b$ both have remainder $r$ and $c$ and $d$ both have remainder $s$ then $a+c$ and $d + d$ will both have whatever remainder $r+s$ is. So the integers $a+c$, $b+d$, and $r+s$ will all belong to the same class of integers.
Likewise $a = q*n + r; c = p*n + r$ and $c = v*n + s; d=w*n + s$ then $ac = qv*n^2 + rvn + sqn + rs$ and $bd = pwn^2 + rwn + spn + rs$. And if $rs = \gamma n + \beta$ then $ac$, $bc$, and $rs$ all belong to the class that has $\beta$ as remainders.
And this is IMPORTANT, there are only $n$ possible classes and every integer is in one.
So to solve the problem:
There are only $7$ possible classes that $n^2$ belong to. And which class each $n$ belongs to will be determined distinctly by what class $n$ is in.
So all you have to do is test each of the seven classes directly.
If $n$ is in the class of integers that has $0$ as a remainder, then $n^2$ will be in the class that has $0$ as a remainder.
If $n$ is in the class that has $1$ as a remainder, then $n^2$ will be in the class that has $1$ as a remainder.
If $n$ is in the class that has $5$ as a remainder. Then $n^2$ is in the same class that $25$ is in. And that is the class that has $4$ as a remainder.
There are only $7$ possibilities and we can check each one.
Now $-3$ is in the same class as $4$ so we could check $-3$ instead of $4$. THis is useful for checking the integers that are in the class that has $6$ as a remainder. They are in the same class that $-1$ is in. So their squares will be in the class of those that have $1$ as a remainder. That is easier than recognizing the will all be in the same class as $6*6 =36$ which is in the class that has $1$ as a remainder.
