Does $\sqrt{1+\sqrt{2}}$ belong to $\mathbb{Q}(\sqrt{2})$?

The truth is that $\sqrt{1+\sqrt2}$ cannot be expressed as $a+b\sqrt2$ where $a,b$ are integers or rationals. If they were reals then the problem becomes trivial.

(Another way to see the first fact above is that the minimal polynomial of $\sqrt{1+\sqrt2}$ is degree-$4$, whereas if it were expressible as $a+b\sqrt2$ with $a,b\in\mathbb Q$ the minimal polynomial would only be quadratic.)

Your start of the question Find $\sqrt{1+ \sqrt 2}$ is misleading.

You're supposing that $\alpha = \sqrt{1+ \sqrt 2}$ belongs to the field extension $\mathbb Q(\sqrt 2) / \mathbb Q$... And you proved that it's not the case.

In fact $\alpha$ is an element of degree $2$ over $\mathbb Q(\sqrt 2)$ because $\alpha$ is a root of the polynomial $p \in \mathbb Q(\sqrt 2) [x]$

$$p(x) = x^2 - (1 +\sqrt 2).$$

And $p$ is irreducible over $\mathbb Q(\sqrt 2)$.