if $x \equiv y$ (mod $m$), then $ax^2 + bx + c \equiv ay^2 + by + c$ (mod $m$) By assumption $y = x + km$ ; some $k \in \mathbb{Z}$. 
w.t.s  $ay^2 + by + c = (ax^2 + bx + c$) + $k'm$ ; some $k' \in \mathbb{Z}$ 
If we substitute our assumption in for $y$ we have 
$a(x + km)^2 + b(x + km) + c$ 
= $a(x^2 + 2xkm + k^2m^2) + bx + c + (bk)m$ 
= $ax^2 + 2xkma + ak^2m^2 + bx + c + (bk)m$ 
= $ax^2 + bx + c + (2xka + k^2ma + bk)m$ 
Lastly, set $k' = 2xka + k^2ma + bk$ and we have 
$ay^2 + by + c = (ax^2 + bx + c$) + $k'm$
does this work? heavily realying just on closure of $\mathbb{Z}$ under + and $\cdot$ 
(I'm fairly new to number theory)
Also, if this has been asked please just refer me to the Paige, I couldn't finds it. I apologize in advance if it has been. 
 A: Your proof is correct, but in a sense it misses the point of the exercise. Here is a better proof:
Assume that $x \equiv y \mod m$. Then, $x^2 \equiv y^2 \mod m$ (since we can take a congruence to the $2$-nd power). Thus, $ax^2 \equiv ay^2 \mod m$ (since we can multiply a congruence by an integer constant -- in this case, $a$). Also, $ax \equiv ay \mod m$ (here, we have multiplied the original congruence by $a$). Finally, $a \equiv a \mod m$. Now, adding the last three congruences together, we obtain $ax^2 + bx + c \equiv ay^2 + by + c \mod m$. Done.
Why do I call this proof "better"? Well, try applying your idea to show that $\left(ax^4+b\right)\left(c-dx^7\left(x+e\right)\right) \equiv \left(ay^4+b\right)\left(c-dy^7\left(y+e\right)\right) \mod m$ instead :) My proof, on the other hand, would still apply here, except that you would have to multiply and subtract congruences as well. The general principle at hand here is the principle of substitutivity for congruences (see §2.2.7 of my Notes on the combinatorial fundamentals of algebra, 10th of January 2019, for example).
A: You may just write
$$a(x^2-y^2) + b(x-y) \equiv (x-y)(a(x+y) +b) \equiv 0 \mod m$$
A: Prove things once.  Then assume they are written in stone.
Prove once that if $a\equiv a'\pmod n$ and $b\equiv b'\pmod n$ then $a \pm b\equiv a' \pm b'\pmod n$.
Then prove once that if $a \equiv a' \pmod n$ and $b\equiv b'\pmod n$ then $ab \equiv a'b'\pmod n$.
Then observe once by induction that if $a \equiv a'$ then $a^m \equiv a'^m \pmod n$ for all positive powers $m$.
And observe once for any integer $d$ that $d \equiv d \pmod n$ for any integer $n$.
Then this is ... just observation.

$x \equiv y \pmod m$ so $x^2 \equiv y^2$ and as $a \equiv a \pmod m$ then $ax^2 \equiv ay^2$ and as $b \equiv b$ and $c \equiv c$ then $bx \equiv by$ and $ax^2 + bx + c \equiv ax^2 + bx + c\pmod n$.

And that IS a complete proof.
=====
To prove these things ONCE:
$c - c = 0$ and $0 = 0*n$ for all integers $n$ so $n|c-c$ for all $n$ so

$c \equiv c \pmod n$ for any integers $c, n$.

If $n|a$ then there is an integer $k$ so that $a = kn$ and if $n|b$ then there is a $j$ so that $b = jn$ so $a\pm b = kn \pm jn = (k\pm j)n$.  $k\pm j$ are both integers so

If $n|a$ and $n|b$ then $n| a\pm b$.

If $a\equiv a' \pmod n$ and $b\equiv b' \pmod n$ then $n|a-a'$ and $n|b-b'$ so $n|(a-a') \pm (b - b')=(a\pm b)-(a' \pm b')$ so

If $a \equiv a'\pmod n$ and $b \equiv b'\pmod n$ then $a\pm b \equiv a' \pm b'\pmod n$.

If $a\equiv a' \pmod n$ then there is a $k$ so that $a = a' + kn$ and if $b\equiv b' \pmod n$ then there is a $j$ so that $b = b' + jn$ so $ab = a'b' + ajn + bkn + jkn^2 = a'b' + n(aj + bk + jkn)$ and as $aj + bk + jkn$ is an integer then

If $a \equiv a'\pmod n$ and $b \equiv b'\pmod n$ then $ab \equiv a'b' \pmod n$.

If we have $a \equiv a' \pmod n$ and we have $a^k \equiv a'^k \pmod n$ for $k = 1$.  If we assume that for some $k$ that $a^k \equiv a'^k$ then we have $a^{k+1} = a^ka \equiv a'^ka'= a'^{k+1} \pmod n$
So by induction $a^m \equiv a'^m\pmod n$ for all positive integer powers $m$.
....
And that's it.  We have proven everything we need once.  We will NEVER need to prove them again.
A: I got it myself. 
Since $x \equiv y$ (mod $m$), by property of $\equiv$ one has
$x^2 \equiv y^2$ (mod $m$) 
By scalar property one has 
$ax^2 \equiv ay^2$ (mod $m$)
Also by scalar and our assumption one has 
$bx \equiv by$ (mod $m$) 
By additive property we have 
$ax^2 + bx \equiv ay^2 + by$ (mod $m$) (**)
Note by using our assumption we also have 
$x + c \equiv y + c$ (mod $m$) 
(Because $m \vert (x - y) \Rightarrow m \vert x + c - y - c$), applying this to (**) and get
$ax^2 + bx + c \equiv ay^2 + by + c$ (mod $m$) and we are done. 
