Consider all the smooth K3 surfaces given by $X^4+W^2X^2+XW^3 = f(Y,Z,W)$ or $X^4+XW^3 = g(Y,Z,W)$ over $\mathbb F_{2}$ with $f$ or $g$ homogenous of degree 4. There are a lot of choices for $f$ and $g$ which give smooth K3 surfaces, for example $f(Y,Z,W) = Y^4 + Y^3Z + YZ^3 + Y^2ZW + YZW^2$.

I've computed several examples and for all of them the traces are either $0$ or a power of two (up to sign). Moreover, the characteristic polynomial of the Frobenius over $\mathbb F_2$ decomposes as a product of cyclotomic polynomials. Hence if the Tate conjecture is true for characteristic 2, then these K3 surfaces are all supersingular.

So the question is, how does one show that they are supersingular? I suppose it comes from the fact that all supersingular K3 surfaces are double covers of the projective plane.

Comment about the computation, if one considers $\mathbb F_{2^q}$ as a linear space over $\mathbb F_2$ then $X \mapsto X^4+X^2+X$ is $\mathbb F_2$ linear and this allows to speed up the computations. Moreover, if $3\nmid q$ then this map is invertible so the number of points over $\mathbb F_{2^q}$ is precisely $|\mathbb{P}^3(\mathbb F_{2^q})|-1$.

Consider all the smooth K3 surfaces given by $X^4+W^2X^2+XW^3 = f(Y,Z,W)$ or $X^4+XW^3 = g(Y,Z,W)$ over $\mathbb F_{2}$ with $f$ or $g$ homogenous of degree 4. There are a lot of choices for $f$ and $g$ which give smooth K3 surfaces, for example $f(Y,Z,W) = Y^4 + Y^3Z + YZ^3 + Y^2ZW + YZW^2$.

I've computed several examples and for all of them the traces are either $0$ or a power of two (up to sign). Moreover, the characteristic polynomial of the Frobenius over $\mathbb F_2$ decomposes as a product of cyclotomic polynomials. Hence if the Tate conjecture is true for characteristic 2, then these K3 surfaces are all supersingular.

So the question is, how does one show that they are supersingular? I suppose it comes from the fact that all supersingular K3 surfaces are double covers of the projective plane.

Comment about the computation, if one considers $\mathbb F_{2^q}$ as a linear space over $\mathbb F_2$ then $X \mapsto X^4+X^2+X$ is $\mathbb F_2$ linear and this allows to speed up the computations. Moreover, if $3\nmid q$ then this map is invertible so the number of points over $\mathbb F_{2^q}$ is precisely $|\mathbb{P}^2(\mathbb F_{2^q})|$.