How come $ax^2+ ty^2 = z^2$ has no solution in $\mathbb{Q}_p$?

Question

Let $a \in \mathbb{Q}^*$ be a non-square and let $p>2$ be a prime. Suppose $a \not \in (\mathbb{Q}_p^*)^2$(i.e. not a square in $\mathbb{Q}_p^*$) and the valuation of $t \in \mathbb{Q}$ with respect to $p$ is $1$. I would like to show that $ax^2+ ty^2 = z^2$ has no solution in $\mathbb{Q}_p$. Any comments would be appreciated. Thank you.