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May 20, 2018 at 15:58 comment added Noam D. Elkies For isotrivial fibrations, there's at least $Y^2 = X^3 - (t^3-t)^2 X$, or more generally $y^2 = Q(t)^3 P(x/Q(t))$ where $u^2 = P(t)$ and $u^2 = Q(t)$ are supersingular elliptic curves in characteristic $p$. These K3 surfaces must all be isomorphic with the Fermat quartic, because the $e=1$ surface is unique; but finding explicit isomorphisms can be tricky.
May 20, 2018 at 14:05 comment added Noam D. Elkies For $p \equiv 3 \bmod 4$ this K3 surface is the "supersingular" surface of Artin invariant $1$; elliptic fibrations correspond to even lattices of rank $20$ with $L^*/L \cong ({\bf Z}/p{\bf Z})^2$, which is a "full classification" of sorts but the number of such lattices grows rapidly with $p$ (there are $52$ for $p=3$, and I recently counted $2683$ for $p=7$).
May 20, 2018 at 12:53 history asked Dimitri Koshelev CC BY-SA 4.0