Rational approximations of $\sqrt{2}$ in $\mathbb{R} \times \mathbb{Q}_7$

Squares of elements of the $3$-adic integers $\mathbb{Z}_3$ are congruent to $0$ or $1$ modulo $3$, thus they are all at $3$-adic distance $1$ from $2$.

Squares of elements of $\mathbb{Q}_3 \setminus \mathbb{Z}_3$ are $3$-adically even further away from the target.

Thus in $\mathbb{Q}_3$, you cannot approximate a square root of $2$ at all well - just like you can't approximate a square root of $-1$ at all well in the reals.

Here is a full solution for the modified problem, inspired by Gro-Tsen's valuable comment.

1. There are infinitely many rational numbers $a/b\in\mathbb{Q}$ in lowest terms such that $$ \left|\frac{a^2}{b^2}-2\right|_\infty\ll\frac{1}{b}\qquad\text{and}\qquad \left|\frac{a^2}{b^2}-2\right|_7\ll\frac{1}{b}.$$ To see this, we work in $\mathbb{Z}[\sqrt{2}]$, the ring of integers of $\mathbb{Q}(\sqrt{2})$. In this ring, the rational prime $(7)$ splits as $(3+\sqrt{2})(3-\sqrt{2})$, while the totally positive units are $(3+2\sqrt{2})^m$. For any positive integer $n$, we can choose the positive integer $m$ so that \begin{align*}a+b\sqrt{2}&=(3+2\sqrt{2})^m(3+\sqrt{2})^n\asymp 7^n,\\ a-b\sqrt{2}&=(3-2\sqrt{2})^m(3-\sqrt{2})^n\asymp 1.\end{align*} This is because $a^2-2b^2=7^n$ holds regardless of $m$. By basic arithmetic in $\mathbb{Z}[\sqrt{2}]$, the integers $a$ and $b$ are relatively prime to each other and to $7$ as well. Moreover, $a$ and $b$ are positive and of size $\asymp 7^n$. It follows that $$\left|\frac{a^2}{b^2}-2\right|_\infty=\frac{7^n}{b^2}\asymp\frac{1}{b} \qquad\text{and}\qquad \left|\frac{a^2}{b^2}-2\right|_7=7^{-n}\asymp\frac{1}{b}.$$

2. By modifying the above argument, we can achieve that $$ \left|\frac{a^2}{b^2}-2\right|_\infty\ll\frac{c}{b}\qquad\text{and}\qquad \left|\frac{a^2}{b^2}-2\right|_7\ll\frac{c^{-1}}{b}$$ for any constant $c>0$. This is essentially best possible, because it is straightforward to see that $$ \left|\frac{a^2}{b^2}-2\right|_\infty\left|\frac{a^2}{b^2}-2\right|_7\geq\frac{1}{|b^2|_\infty|b^2|_7}\geq\frac{1}{b^2}.$$