In my lecture notes there is the following example:
The diophantine equation $y^2=x^3+7$ has no solutions.
Proof:
If the equation would have a solution, let $(x_0, y_0)$, $y_0^2=x_0^3+7$, then $x_0$ is odd.
(If $x_0$ is even, $x_0=2k \Rightarrow x_0^3 \equiv 0 \mod 8$, so $x_0^3+7\equiv 7 \pmod 8$
$\Rightarrow y_0^2\equiv 7 \pmod 8$, contradiction, since $y_0^2 \equiv 0, 1, 4 \pmod 8$ )
$$y_0^2=x_0^3+7 \Rightarrow y_0^2+1=x_0^3+8=(x_0+2)(x_0^2-2x_0+4)=(x_0+2)[(x_0-1)^2+3]$$
$(x_0-1)^2+3 \in \mathbb{N}$
$(x_0-1)^2+3>1$
It stands that $(x_0-1)^2+3 \equiv 3 \pmod 4$.
There is at least one prime divisor of $(x_0-1)^2+3$, of the form $p \equiv 3 \pmod 4$, that means that $$(x_0-1)^2+3 \equiv 0 \pmod p , \text{ where } p \equiv 3 \pmod 4 \\ =y_0^2+1 \equiv 0 \pmod p \\ \Rightarrow y_0^2 \equiv -1 \pmod p$$
$$Y^2 \equiv -1 \pmod p \text{ has a solution } \Leftrightarrow \left ( -\frac{1}{p} \right )=1 \Leftrightarrow (-1)^{\frac{p-1}{2}}=1 \Leftrightarrow p \equiv 1 \pmod 4$$
So $y_0^2 \equiv -1 \pmod p$ doesn't have a solution.
$$$$
Can you explain to me why we check at $\pmod 8$ if $x$ is odd or even???
And do we look after that n $\pmod 4$ ???
The motivation of using mod 8 in the first step is that if $x$ is even $x = 2k$, then $x^3 = 8k^2$ so $x^3$ is zero mod 8; and then you can see that you are solving for $y^2 = -1 \mod 8$ which looks like it might have no solution (as is in fact the case). All the author was trying to do there is to show that $x$ can't be even.
The motivation for the mod 4 is because you are working with $(x_0-1)^2$ and you know that $x_0$ is odd so $(x_0-1)$ is even and its square is divisible by 4.
The last point that might be unclear is why, when you know that $(x_0-1)^2+3$ is of the form $4n+3$, you also know that it has a prime divisor of the form $4m+3$. But that is easy: even prime factors don't matter, and any product of factors of the form $4s_i + 1$ is itself of the form $4n+1$, so to get $4n+3$ you must start from at least one factor of that form.