I want to classify, up to isomorphism, the commutative unital $\mathbb{F}_2$-algebras $A$ for which $a^2\in\lbrace 0,1\rbrace$ for all $a\in A$. So far I have the following:
I have shown that there exists $x,y,z,w\in A$ with $x^2=y^2=z^2=w^2=0$ and for which $\lbrace 1,x,y,z,w\rbrace$ is a basis. If all pairs of products $xy,yz$ etc. are zero, then we have the exterior algebra on $x,y,z,w$ mod these products.
Otherwise, there must exist a pair of generators with non-zero product. WLOG assume $xy\neq 0$. Then $\lbrace 1,x,y,xy\rbrace$ is linearly independent, and there exists $p\in A$ with $p^2=0$ for which $\lbrace 1,x,y,p,xy\rbrace$ is a basis. It follows from this that $xyp=0$, and that $yp,px\in\lbrace 0,xy\rbrace$. Since we can interchange $x$ and $y$ by symmetry, the cases $yp=0,xp=xy$ and $yp=xy,xp=0$ are isomorphic.
Relabeling $p$ as $z$, we have the exterior algebra on $x,y,z$ mod $yz,zx$ as case 1, mod $yz,xy+xz$ as case 2, and mod $xy+yz,xy+xz$ as case 3.
However, I suspect that the case of three generators should force $yz,zx$ to be zero. More generally, I am trying to construct an $\mathbb{F}_2$-algebra with the above property of countable dimension suitably containing all the finite ones. Have I gone wrong somewhere?
Adding the following comment that won't fit into one, so an answer it is.
Let $I$ be the 4-dimensional kernel of the squaring homomorphism (of rings) from $A$ to $\Bbb{F}_2$. As you observed, $I$ is an ideal. The elements outside of $I$ have squares equal to $1$, so $I$ is the unique maximal ideal of $A$ consisting of the non-units (so $A$ is local).
With $x,y\in I$ consider the mappings $f_x:A\to A, a\mapsto xa$ and $f_y:A\to A, a\mapsto ya$. Because $f_x\circ f_x=0$, rank-nullity forces $\dim \mathrm{Ker}(f_x)\ge3$, and similarly $\dim \mathrm{Ker}(f_y)\ge3$.
Assume that $x\in I$ and $y\in I$ are as in your argument, so $xy\neq0$. In that case both kernels have dimensions exactly three. As ideals they are both subspaces of $I$, so the two kernels must intersect in a subspace $W$ of dimension two. But $W\cap \langle 1,x,y,xy\rangle$ is obviously spanned by $xy$. Therefore there exists another element $z\in I$, linearly independent from $xy$, annihilated by both $x$ and $y$.
So without loss of generality you can assume that $zx=zy=0$. I think this should let you proceed with the classification.