How many nonisomorphic associative algebras of dimension 2 over the field F_{p^k} are there? As much as I have searched, I have not found any results that answer my question; not even for k = 1,2.

$\begingroup$ See maa.org/sites/default/files/… for rings of order $p^2$ but warning the rings are not assumed unital $\endgroup$– Benjamin SteinbergCommented Feb 10, 2021 at 15:20

$\begingroup$ Probably it makes sense to ask the question over an arbitrary (perfect?) field and then specify. Also it's reasonable to first ask about classification modulo being geometrically isomorphic. $\endgroup$– YCorCommented Feb 11, 2021 at 16:19
2 Answers
The answer is: 8 isomorphism classes. The classification up to absolute isomorphism yields: 7.
To show this in a greater and more natural generality, let in general $K$ be a field. I claim that there are 7 isomorphism classes of 2dimensional $K$algebras that are not (commutative) fields. Denote by $0_i$ the $i$dimensional vector space endowed with null product. The isomorphism classes are represented by the $K$algebras with basis $(x,y)$ for which the nonzero products among $xx,xy,yx,yy$ are described by
$W_1$: $xx=x$, $xy=y$;
$W_2$: opposite to $W_1$ ($xx=x$, $yx=y$);
(the remaining ones are commutative)
$W_3$: $0_2$;
$W_4$: $K\times 0_1$: $xx=x$;
$W_5$: $xx=y$;
(the remaining ones are unitary)
$W_6$: $xx=x$, $xy=yx=y$ ($K[t]/t^2$);
$W_7$: $xx=x$, $yy=y$ ($K\times K$).
This being granted (see below for a proof), it remains to classify quadratic extensions of $K$. If $K$ has characteristic $\neq 2$, they are classified by $K^*/K^{*2}\smallsetminus\{1\}$, which has order 2 when $K$ is finite. [This corresponds to twisted forms of $W_7$, if we include $W_7$ itself].
If $K$ has characteristic $2$, the separable quadratic extensions are classified by the cokernel of the additive endomorphism $t\mapsto t+t^2$ of $K$ (minus $\{0\}$), and again this cokernel has order 2, and again, including $W_7$, this are forms of $W_7$.
This covers finite fields. If we are interested in nonperfect fields of characteristic 2, we get the inseparable quadratic extensions, which are twisted forms of $W_6$. Including $W_6$, these are classified by the orbits of $K^{*2}\rtimes K^{*2}$ on $K$, acting by homotheties.
It remains to prove the classification above $W_{1\le i\le 7}$. Let $A$ be a 2dimensional associative algebra.
First, the commutative unital case, which is the most standard (already covered by the previous answer): being artininan, if not local, $A$ is a product of at least 2 local unital commutative algebras, and the only possibility is $W_7$. If local and not a field, the maximal ideal $M$ has dimension 1, $M^2=0$ and we get $W_6$.
Second assume it commutative, but not unital. Adding a unit realizes our algebra $A$ as ideal of dimension $3$ in a 3dimensional unital artinian algebra $A'$. If $A'$ is a product of fields, all its ideals are unital (as algebras, albeit not unital subalgebras). So the decomposition of $A'$ as product of local algebras involves a nonfield. So either $A'$ is local, or $A'\simeq K\times K[t]/t^2$. In the second case, the two maximal ideals of $A'$ are $\{0\}\times K[t]/t^2$ (which is unital), and to $W_4$. Next suppose $A'$ is local. If $x^2=0$ for all $x\in A$, then in characteristic not $2$, we deduce $xy=0$ for all $x,y$ by polarization, and we have $A\simeq W_3$. If $x^2\neq 0$, then setting $y=x^2$ we see that $A\simeq W_5$.
(In char. 2 it remains to exclude another possibility, where $x^2=0$ for all $x$. If the product were nonzero, let $z$ be in its image, complete to a basis and rescale to get $xx=zz=0$, $xz=zx=z$, and then $x(xz)(xx)z=xz0=z$, contradicting associativity.)
Third and last, assume $A$ not commutative. There's a single noncommutative 2dimensional Lie algebra up to isomorphism, namely with basis $(x,y)$ such that $[x,y]=y$. Hence (applying this to the associated Lie commutator) we can suppose that $A$ has a basis $(x,y)$ with $xyyx=y$. Since $A$ is noncommutative and has dimension $\le 2$, the centralizer of every nonzero element is a proper subspace, and hence (using associativity), $x^2$ is a scalar multiple of $x$ for every $x$. So, we can write $xx=ax$, $yy=by$, $yx=cx+dy$, $xy=cx+(d+1)y$.
Then (u) $0=y(yx)(yy)x=y(cx+dy)(by)x=c(cx+dy)+dbyb(cx+dy)=c((cb)x+dy)$.
Assume by contradiction $c\neq 0$. By (a) we deduce $d=cb=0$. Thus $xx=ax$, $yy=cx$, $xy=cx+y$, $yx=cx$. Then $0=(xy)xx(yx)=(cx+y)xx(cx)=yx$, so $cx=0$, contradiction. Hence $c=0$, contradiction.
So $c=0$: $xx=ax$, $yy=by$, $yx=dy$, $xy=(d+1)y$.
Then $0=y(xy)(yx)y=((d+1)d)y^2=y^2=by$, so $b=0$.
Next $0=(yx)xy(x^2)=dyxy(ax)=(da)dy$. So $(da)d=0$. Also $0=x(xy)(x^2)y=(d+1)xyaxy=(d+1a)(d+1)y$. So $(d+1a)(d+1)=0$.
If $d=0$, the latter writes $(1a)=0$, so $a=1$, and $A\simeq W_1$. Otherwise $d=a$, and the latter writes $d+1=0$, so $d=1=a$. In this case $A\simeq W_2$ (change $x$ into $x$ to get the defining basis of $W_2$).
Finally the $W_i$ are pairwise nonisomorphic: we see that $W_1$ and $W_2$ are not isomorphic because $x$ is a left unit in $W_1$ while $W_2$ has no left unit. The other cases are straightforward.
PS I don't know if there's a reference for this. However, the complex and real cases are easy to locate in the literature. From the complex case (7 cases), it formally follows that the same classification holds in large enough characteristic for algebraically closed fields. Moreover in all cases $W_i$ has an automorphism group whose automorphism group has no Galois cohomology, so one can predict beforehand that there's a single form over any perfect subfield. The exception being $K\times K$, whose forms correspond to quadratic separable extensions. Anyway I preferred not invoke Galois cohomology in the arguments above.
Assuming that your algebra $A$ is unital, with $1\in A$, and that $A$ is an $F=GF(p^k)$algebra, meaning that $F=F\cdot 1 \leq A$ is a central subalgebra, then $A=F\oplus F\: \overline{x}$ for some $\overline{x}\in A$. In particular $A$ is commutative and $F[x]\rightarrow A$ with $x\mapsto \overline{x}$, an surjective homomorphism of algebras, and $A\simeq F[x]/\langle p(x) \rangle$ for some monic polynomial $p(x)\in F[x]$ of degree $2$. Then either
 $p(x)$ is irreducible and $A$ is a field with $p^{2k}$ elements, which up to isomorphism is unique, or
 $p(x)$ has two distinct roots, say $a$ and $b$, hence $A\simeq F[x]/\langle xa\rangle \times F[x]/\langle xb\rangle \simeq F\times F$ or
 $p(x)$ has only one root $a$ of multiplicity $2$ and $A\simeq F[x]/\langle (xa)^2\rangle \simeq F[x]/\langle x^2\rangle$.
So there should be only three nonisomorphic algebras of dimension $2$. The situation is probably quite different, if $A$ is nonunital.