The following paper gives a large family of K3 surfaces for which the authors prove that the geometric Picard number is 3. The generators are quite explicit, since the surfaces are defined by $(2,2,2)$ forms on $\mathbb P^1\times\mathbb P^1\times\mathbb P^1$.

Baragar, Arthur; van Luijk, Ronald;
K3 surfaces with Picard number three and canonical vector heights.
*Math. Comp.* **76** (2007), no. 259, 1493–1498.  MR2299785

EDIT: Sorry, just realized that you want quartic K3's sitting in $\mathbb P^3$, not K3's in  $\mathbb P^1\times\mathbb P^1\times\mathbb P^1$. But maybe this paper will still be of interest, so I won't delete this answer. In any case, based on the first three answers, it seems that the correct answer is "ask van Luijk", since he's the leading expert in proving this sort of thing.