Using Hensel's lemma, it's standard fair to show that if $p$ is odd, $d\in\mathbb Z$, and $(p,d) = 1$, then $\sqrt{d}\in\mathbb Q_p$ if and only $d$ is a quadratic residue mod $p$. Naturally, there are two follow-up questions to this.
(1) If $p$ is odd and $p|d$, when is $\sqrt{d}\in\mathbb Q_p$?
(2) Letting $f(x) = x^2-d$, since $f'(x) = 2x \equiv 0\bmod{2}$, so how do we handle the case of $p = 2$? When $d\in\mathbb Z_2^\times$, this is answered in Theorem 4.4 of Professor Conrad's notes. This leaves the case of $d\notin\mathbb Z_2^\times$ just as in (1).
My progress on is as follows: writing $d = p^r m$ with $r>0$, if $r$ is even then $\sqrt{d}\in \mathbb Q_p$ if and only if the same holds for $n$. The remaining cause is $r$ odd in which case factoring $p^{r-1}$ out, the question reduces to
(1') When is $\sqrt{pm}\in\mathbb Q_p$ when $(p,m)=1$?
Is there anything intelligible that can be said here?
Note that if $\sqrt{d}$ is in $\Bbb Q_p$, i.e. $d = x^2$ for some $x\in\Bbb Q_p$, then the valuation $v_p(d)$ is equal to $2v_p(x)$, hence must be an even integer, as $v_p(x)$ is an integer.
This means that in your "remaining case" where $d=p^rm$ with $r$ odd, the number $d$ cannot be a square in $\Bbb Q_p$.
The rest of your arguments are all correct.