Find primes $p$, for which the $x^2 \equiv 7 \pmod{p}$ has a solution 
I want to determine for which primes $p$ the following congruence has a solution.
$$x^2 \equiv 7 \pmod{p}.$$

I have thought the following.
We are looking for the primes $p$ for which the Legendre symbol $\left( \frac{7}{p}\right)=1$. Then $(7,p)=1$.
If $p$ is odd, we have from Euler's theorem that
$$\left( \frac{7}{p}\right)=7^{\frac{p-1}{2}} \pmod{p}$$
So we would want that $7^{\frac{p-1}{2}} \pmod{p}=1$.
But does this help in order to find the desired primes $p$ ?

It's a little longer. From law of quadratic reciprocity, we have
$$\biggl(\frac7p\biggr)=\biggl(\frac p7\biggr)(-1)^{\tfrac{p-1}2},$$ 
\begin{align}&\text{On the other hand, }&&\begin{cases}(-1)^{\tfrac{p-1}2}=1&\text{if }\;p\equiv 1\mod 4,\\(-1)^{\tfrac{p-1}2}=-1&\text{if }\;p\equiv 3\;(\text{or }-1)\mod 4.
\end{cases}&&\hspace{10em}\end{align}
Also the list of non-zero squares mod. $7$ is$\enspace\bigl\{1,\,2,\,4\bigr\}$. 
So we have to solve the systems of congruences:
\begin{align}
&\begin{cases}p\equiv 1\mod4,\\p\equiv 1,\,2,\,4\mod 7,
\end{cases}
& &\qquad\begin{cases}p\equiv -1\mod4,\\p\equiv 3,\,5,\,6\mod 7,
\end{cases}
\end{align}
To solve these systems, we use the explicit formula for the inverse isomorphism of the abstract form of the Chinese remainder theorem. 
More precisely, start from a Bézout's relation between $4$ and $7$, e.g. 
$\;4\cdot 4-1\cdot 7=1$. Then
$$\begin{cases}p\equiv a\mod4,\\p\equiv b\mod 7
\end{cases}\iff p\equiv  4\cdot 4\, b- 7a \mod 28$$
In all, you should find, if I'm not mistaken:
$$p\equiv \pm 1,\pm3,\pm 9\mod 28. $$
Added:
As @OscarLanzi observed, this list can be simplified to
$$x^2 \equiv 7 \;\text{ has  a solution mod. }p \iff p \equiv 3^k \mod 28 \;\text{ for some }k.$$
This is  because $3$ has order $6\bmod 28$. Indeed the successive power of $3$ mod. $28$ are:
$$ 1,\;3,\;9,\;3^3\equiv -1,\;3^4\equiv -3,\;3^5\equiv -9, \;3^6\equiv -27\equiv 1 \mod 28.$$
