Just take all polynomials in your nilpotent element $x$ (the generator of $Z$ as a $K$ algebra), with constant term zero. This is clearly an ideal of $Z$, say $m$. Note that $m$ is a proper ideal because $1$ is not in $m$.
Now an element not in $m$ is a polynomial $a_nx^n + ... + a_1x + a_0$, where $a_i \in K$, and $a_0 \neq 0$, so a unit in $Z$. But, $a_nx^n + ... + a_1x$ is nilpotent being a sum of nilpotent elements. If $a_nx^n + ... + a_1x + a_0$ is not a unit, then there is some max ideal $m'$ such that $a_nx^n + ... + a_1x + a_0 \in m'$. But, nilpotents are in every max ideal. So, $a_nx^n + ... + a_1x \in m'$, which shows that $a_0 \in m'$, which is impossible because $a_0$ is a unit. Thus, $a_nx^n + ... + a_1x + a_0$ must be a unit. Hence, any element of $Z/m$ is a unit, and $m$ is the only maximal ideal.