Problem: Prove that $(x+4)^2 ≡ x^2 \pmod 8$ , for all $x \in \mathbb{Z}$
Since the definition of congruence is: If $a-b$ are divisible to $n$, $a$ and $b$ are congruent modulo $n$, $a \equiv b \pmod n$
$$ \begin{align*}\implies &(x+4)^2 - x^2 \over 8 &= k &\quad \text{where,}\space k\in \mathbb{Z} \\ \implies &x^2 + 4x + 4x + 16 - x^2 \over 8 &= k \\ \implies &8x + 16 \over 8 &= k \end{align*}$$
$\implies$ Since both $8x$ and $16$ are multiple of $8$, their sum of them is also divisible by $8$ for every integer $x$.
Is this a correct proof?
An easy way:
$$(x+4)^2\equiv x^2+8x+16\equiv x^2\mod 8$$
But yes, your approach is correct (in fact, it's equivalent to mine). However, you should write it the other way around: in a formal proof, you should start with what you know and work towards what you want. You work from what you want to what you know (which is fine, and usually finding a way to prove something is easier this way, but keep everything you do reversible). So if you were to write a nice, formal, proof, I'd do something like this:
Let $k=x+2$, so that $k$ is an integer for every $x\in\Bbb Z$. Now we see $k=\frac{8x+16}{8}$, which we can also write as
$$k=\frac{x^2+8x+16-x^2}{8}=\frac{(x+4)^2-x^2}{8}$$
This means that $8$ is a divisor of $(x+4)^2-x^2$, which we can write as $$(x+4)^2\equiv x^2\mod 8$$
and this proves the statement we wanted. $\square$