Ring of order $p^2$ must be commutative. See this relevant post
Ring of order $p^2$ is commutative.
Either I don't understand the above or it isn't quite complete (presumably the former).
Let $R$ be a ring of order $p^2$. Then $char(R)=p$ and $Z(x)=p,p^2$ for any $x \in R$. In the latter case there is nothing to show. Hence, let $Z(x)=p$. Then $Z(x) \cong \mathbb{Z}/p\mathbb{Z}$ (as a group not necessarily a ring) and supposedly this is enough to say that $R$ is commutative but I do not readily see this. Especially since ring homomorphisms must send $1 \to 1$. A detailed explanation would be appreciated. 

We recall that $R$ has an additive subgroup of order $\text{char}(R)$; thus, since $\vert R \vert = p^2$, either $\text{char}(R) = p$ or $\text{char}(R) = p^2$.
If 
$\text{char}(R) = \vert R \vert = p^2, \tag 1$
we are done, since every element of $r$ is a sum of $1$s; so suppose
$\text{char}(R) = p; \tag 2$
then
$\Bbb Z_p \simeq \Bbb F_p \subsetneq R, \tag 3$
that is, $\Bbb F_p$ is a sub-field of $R$, isomorphic to $\Bbb Z_p \simeq \Bbb Z/p\Bbb Z$; thus, $R$ is an $\Bbb F_p$-vector space; now let
$R \ni x \notin \Bbb F_p; \tag 4$
such an $x$ exists since $\vert R \vert = p^2 > p = \vert \Bbb F_p \vert$, so $R \setminus \Bbb F_p \ne \emptyset$.  I claim $1_R$ and $x$ are linearly independent over $\Bbb F_p$; if not, then there are $a, b \in \Bbb F_p$, not both $0$, with
$a + bx = a1_R + bx = 0, \tag 5$
or
$x = -b^{-1}a \in \Bbb F_p, \tag 6$
contrary to hypothesis; we conclude that $1_R$ and $x$ are in fact linearly independent over $\Bbb F_p$;  therefore $1_R$ and $x$ span a $2$-dimensional subspace of $R$; but a $2$-dimensional vector space over a $\Bbb F_p$ has precisely $p^2$ elements.  Therefore $1_R$ and $x$ in fact span $R$; thus given $y, z \in R$ we may write
$y = a + bx, \; z = c + dx, \; a, b, c, d \in \Bbb F_p; \tag 7$
it then follows that
$yx = (a + bx)(c + dx) = ac + (ad + bc)x + bdx^2$
$= ca + (da + cb)x + dbx^2 = (c + dx)(a + bx) = zy, \tag 8$
and we see that $R$ is commutative.
