I think the answer to the question in the title is in the affirmative. But I am not sure how to go about proving or disproving it. Here is what I would like to do.
Suppose $\sqrt{3}$ $\notin$$\mathbb{Q}(\sqrt[4]{2})$. Denoting $\mathbb{Q}(\sqrt[4]{2})$ by $F$, note that $F = L_{F}(\{1\})$ where $L_F(S)$ denotes the linear span of $S$ over $F$. Since $\sqrt{3}$ is not in $L_{F}(\{1\})$, we see that $\{1, \sqrt{3}\}$ is linearly independent (in some vector space over the field $F$, say $\mathbb{R}$). Thus, dimension of $L_F(\{1,\sqrt{3}\})$ is 2. The inclusion $L_F(\{1,\sqrt{3}\})$$\subset$$\mathbb{Q}(\sqrt[4]{2},\sqrt{3})$ is clear. I am not sure about the opposite containment. If I could show it, then it can be proved that $[\mathbb{Q}(\sqrt[4]{2},\sqrt{3}):\mathbb{Q}(\sqrt[4]{2})]=2$.
Please keep the discussion elementary, my knowledge of field theory does not go beyond the very first section of the book, Fields and Galois Theory by Patrick Morandi.