Here's what I've thought so far. Unless I'm very much mistaken, we have that $[\mathbb{Q}(\sqrt{3},\sqrt{5}):\mathbb{Q}] = 15$, so I'm looking for a $\theta$ such that $[\mathbb{Q}(\theta):\mathbb{Q}] = 15$. Because it's supposed to be a simple extension, I need the minimal polynomial of $\theta$ over $\mathbb{Q}$ to be of degree $15$. I'm quite sure $\theta$ cannot be $15$, although I have only briefly sketched out why I think that is.
I'm looking for some hints on how to approach this problem. I'll update this post with an edit once I have "solved" it and hopefully someone will tell me if it's correct or not.
You should be thinking geometrically. The field $K=\Bbb Q(\sqrt3,\sqrt5\,)$ is of degree four over $\Bbb Q$, as the comments have pointed out. There are only four proper subfields, namely $\Bbb Q$, $\Bbb Q(\sqrt3\,)$, $\Bbb Q(\sqrt5\,)$, and $\Bbb Q(\sqrt{15}\,)$. Each of the fields involving a square root is two-dimensional. So here you are, in a four-dimensional space, with three two-dimensional subspaces, all intersecting in the one-dimensional subspace $\Bbb Q$.
You haven’t come anywhere near exhausting all the elements of $K$ — those fields are thin closed subsets of the whole of $K$. Any element $\alpha$ of $K$ that isn’t in the union of those three will have the property that $\Bbb Q(\alpha)=K$.