Prove that $x^{2} \equiv -1 \pmod p$ has no solutions if prime $p \equiv 3 \pmod 4$. Assume: $p$ is a prime that satisfies $p \equiv 3 \pmod 4$

Show: $x^{2} \equiv -1 \pmod p$ has no solutions $\forall x \in \mathbb{Z}$.

I know this problem has something to do with Fermat's Little Theorem, that $a^{p-1} \equiv 1\pmod p$. I tried to do a proof by contradiction, assuming the conclusion and showing some contradiction but just ran into a wall. Any help would be greatly appreciated.
Suppose $x^2\equiv -1\pmod{p}$. Then $x^4\equiv 1\pmod{p}$. Since $p = 4k+3$, we have
$$x^{p-1} = x^{4k+2} = x^2x^{4k} \equiv -1(x^4)^k\equiv -1\pmod{p},$$
which contradicts Fermat's Little Theorem.
A solution to the given congruence would imply there exists an element of multiplicative order 4 in the finite field of size p which is false because (p-1)/2 is odd and therefore p-1 is not divisible by 4.