Is there also an other way, to prove that the diophantine equation has no solution?

I am looking at the following exercise:

The diophantine equation $y^2=x^3+7$ has no solution.

If the equation would have a solution, let $(x_0,y_0)$,then $x_0$ is odd.

( If $x_0$ is even, $x_0=2k \Rightarrow x_0^3 \equiv 0 \pmod 8$, therefore $y^2 \equiv 7 \pmod 8$, that is a contradiction, since $y^2 \equiv 0,1,4 \pmod 8$ )

$$y_0^2=x_0^2+7 \Rightarrow y_0^2+1=x_0^3+8=x_0^3+2^3=(x_0+2)(x_0^2-2x_0+4)=(x_0+2)[(x_0-1)^2+3]$$

$$(x_0-1)^2+3 \in \mathbb{N}$$ $$(x_0-1)^2+3>1 , \text{ so it has a prime divisor, let }p$$

It stands $(x_0-1)^2+3 \equiv 3 \pmod 4$

If $p_1 \equiv 1 \pmod 4$ and $p_2 \equiv 1 \pmod 4$, $p_1 \cdot p_2 \equiv 1 \pmod 4$.

Therefore, there is at least one prime divisor of $(x_0-1)^2+3$, of the form $p \equiv 3 \pmod 4$, so $(x_0-1)^2+3 \equiv 0 \pmod p \Rightarrow y_0^2+1 \equiv 0 \pmod p \Rightarrow y_0^2 \equiv -1 \pmod p$

$$Y^2 \equiv -1 \pmod p \text{ has a solution } \Leftrightarrow \left ( \frac{-1}{p}\right )=1 \Rightarrow (-1)^{\frac{p-1}{2}}=1 \Rightarrow \frac{p-1}{2}=2\\ \Rightarrow p \equiv 1 \pmod 4$$

So, $y_0^2 \equiv -1 \pmod p$ has no solution.

Is there also an other way, to prove that the diophantine equation $y^2=x^3+7$ has no solution?

It so happens that the ring $\mathbb{Z}[\sqrt{7}]$ is a UFD, so we can rewrite this as:
$$(y-\sqrt{7})(y+\sqrt{7}) = x^3$$
If $7 \mid y$, we get a contradiction $\pmod{49}$. Likewise, if $2\mid y$, we get a contradiction $\pmod{8}$ (your own reasoning shows this).
Otherwise, $\gcd(y-\sqrt{7}, y+\sqrt{7}) = 1$, implying that both $y-\sqrt{7}$ and $y+\sqrt{7}$ are cubes in $\mathbb{Z}[\sqrt{7}]$.
So there exist $a,b\in\mathbb{Z}$ with $y+\sqrt{7} = (a+b\sqrt{7})^3 = (a^3 + 21ab^2) + (3a^2 b + 7b^3)\sqrt{7}$, so $y=a^3+21ab$, $1=3a^2b + 7b^3$.
The second equation is $b(3a^2 + 7b^2) = 1$. But $3a^2+7b^2$ is never a divisor of $1$, contradiction.