Show that $\frac {(\mathbb Z/p)[x]}{(x^2)}$ is not isomorphic to $\frac {(\mathbb Z/p)[x]}{(x)} \times \frac {(\mathbb Z/p)[x]}{(x)}$ I am given in the question that $\frac {(\mathbb Z/p)[x]}{(x^2)}$ has only one maximal ideal.

So my thinking for this is that   $(\mathbb Z/p)[x]$ is an integral domain.

Now my original thinking was to use Chinese Remainder Theorem as I thought the $\gcd(x,x) = x$ which means they are not coprime so they can't be isomorphic.

I have a secondary thought of showing that   $\frac {(\mathbb Z/p)[x]}{(x)}$ has one maximal ideal, therefore its product $\frac {(\mathbb Z/p)[x]}{(x)} \times \frac {(\mathbb Z/p)[x]}{(x)}$ will have two maximal ideals, but I am not sure whether this is appropriate too.
Hint:

$\mathbf Z/p\mathbf Z[x]/(x^2)$ has a nonzero nilpotent element, whereas

$\mathbf Z/p\mathbf Z[x]/(x)\times\mathbf Z/p\mathbf Z[x]/(x)\simeq\mathbf Z/p\mathbf Z\times\mathbf Z/p\mathbf Z$, a product of fields.
I would simply show you can't create a bijection that satisfies the properties to make the two isomorphic. $\Bbb{Z}_p [x] \big/ \langle x^2 \rangle$ will have polynomials of a greater order, so you cannot construct said bijection.