In reals, you can use the rule that in real polynomials, you always get roots in conjugate pairs, so a quadratic real polynomial is enough to get any complex number. That means that as long as you can rewrite $\sqrt{i+\sqrt{2}}$ in proper cartesian form or polar form (into a form which you know how to conjugate), you can write down a second order polynomial with that root. I'll use the polar form:
$$z=\sqrt{i+\sqrt{2}}=\sqrt{\sqrt{{1+\sqrt{2}^2}}\exp(i \arctan{(1/\sqrt{2})})}=\sqrt[4]{3}\exp(i \arctan{(1/\sqrt{2})}/2)$$
Now you just have to write down the polynomial
$$(x-z)(x-z^\ast)=0$$
which can be then simplified to
$$x^2-2x \operatorname{Re}(z)+|z|^2=0$$
$$x^2-2x \sqrt[4]{3}\cos\left(\frac{\arctan{2^{-1/2}}}{2}\right)+\sqrt{3}=0$$
The ugly terms are the reason that this doesn't work in rational and extended rational field.
When you have different field, you allow different coefficients, so the polynomials will of course be different unless they are all pure rational (in which case the field extension wasn't needed).
Let's compare $\mathbb{Q}(i)$ and $\mathbb{Q}(\sqrt{2})$ solutions.
For both, you first square it:
$$x^2=i+\sqrt{2}$$
For $\mathbb{Q}(i)$ you need to remove the square root but $i$ is ok:
$$(x^2-i)^2=2$$
$$x^4-2ix^2-3=0$$
For $\mathbb{Q}(\sqrt{2})$ you do the opposite: you remove the $i$:
$$(x^2-\sqrt{2})^2=i^2$$
$$x^4-2\sqrt{2}x^2+3=0$$
These are evidently different. To remove the $i$ or $\sqrt{2}$ to get to pure rational, you will need to square again, getting something of order $8$.