This is essentially the Gelfond-Schneider Theorem which says:
$$
\text{If } a,b \text{ are algebraic}, a \neq 0,1 \text{ and } b \in \mathbb{R} \setminus \mathbb{Q} \text{ then } a^b \text{ is transcendental.}
$$
Now every transcendental number is also irrational, and every rational number is algebraic. $\sqrt{2}$ is also algebraic so in this case, yes $2^\sqrt{2}$ is irrational and transcendental.
But (I believe) a non-zero algebraic number to the power of a transcendental number can either be rational or irrational, so a rational to the power of an irrational may be rational, only if the irrational is also transcendental. For concrete examples note that $\log_2 (3)$ is irrational, but $2^{\log_2 (3)} = 3$, however I suspect $2^\pi$ is irrational.
Examples of algebraic numbers are say $\sqrt{n}$ or $\sqrt{3 + \sqrt{2}}$ or anything that is the root of a polynomial (that is in fact the definition), whereas transcendental numbers are more like $\pi$ or $e$.
Hopefully this was helpful for you, but in that particular example, yes $2^\sqrt{2}$ is irrational.