No, your proof is incorrect because you're assuming that $\left(\dfrac{a}{n}\right)=1$ iff $a$ is a quadratic residue modulo $n$
This is true in case of the Legendre symbol when the denominator is an odd prime but not for the Jacobi symbol.
Quoting from Wikipedia:
But, unlike the Legendre symbol:
If (a/n) = 1 then a may or may not be a quadratic residue modulo n.
The correct way would be to show that $-1$ is a quadratic residue mod $p$ since $\left(\dfrac{-1}p\right)=(-1)^{(p-1)/2}=1$ since $p\equiv 1\pmod4$, thus $(p-1)/2$ is even.
Now, use Hensel's lifting lemma to show that $-1$ is a quadratic residue modulo $p^2$.
You can also generalize this to higher powers of $p$, i.e., $-1$ is a quadratic residue modulo $p^k~\forall~k\geq 2$ using Hensel's lemma (or induction, which mimics the proof of Hensel's lemma, see here)