Let $p,q$ be different primes unequal to $2$. Let $(a/b)$ denote the Legendre Symbol. The following holds:
$q\text{ is a square }\bmod p \Longleftrightarrow (q/p) = 1 \Longleftrightarrow X^2+qY^2 = 0$ has a solution over $\mathbb{Q}^*_p$.
So $X^2 = q + mp$
If i set $Y = 1$
i get $$X^2 = qY^2 + mp.$$
So
$$X^2 - qY^2 = 0 \bmod p$$
holds.
But the sign of $qY^2$ doenst match with my beginning statement. That would mean that
$$(q/p) = 1 \Longleftrightarrow X^2 - qY^2 = 0\text{ has a solution over }\mathbb{Q}^*_p$$
holds.
I am confused. Is my primal statement false?
What you are looking for is (a very weak version of) Hensel's lemma : if $q$ is not divisible by $p$, $X^2 - q$ has a solution in $\mathbb{Q}_p$ if and only if it has a solution in $\mathbb{F}_p$ (this is because $X^2 - q$ is separable in $\mathbb{F}_p$).
So indeed $(q/p)=1$ iff $q$ in a square in $\mathbb{Q}_p$ (which is equivalent to your equation $X^2 - qY^2 = 0$, though using two variables is overkill).