Let $p$ be a prime and $r$ a positive integer. Let $\mathbb Z_{p^r}$ the ring of integers modulo $p^r$. Every zero-divisor $z$ in $\mathbb Z_{p^k}$ can be written in the form $z=ap^k$, where $a$ is a unity.
1) Let now $(R, \mathfrak m)$ be a finite commutative local ring. If the maximal ideal $\mathfrak m$ is principal and generated by $s$, then every zero divisor $z$ of $R$ can also be written in the form $z=as^k$ where $a$ is a unity or we need to impose some conditions on the ring $R$ in order to conclude that $z=as^k$ where $a$ is a unity?
2) Assume now that $(R, \mathfrak m)$ is a finite commutative local ring. Suppose that the maximal ideal $\mathfrak m$ is generated by $s_1, s_2, \dots, s_l$. What conditions we impose to the ring in order to have that a zero-divisor $z$ can be written in the form $z=a_1s_1^{k_1}+\cdots+a_ls_l^{k_l}$, where $a_i$ are a units, for $i=1, \dots , l$?
In (1), the ring $R$ is assumed finite, and so is an Artin local ring with principal maximal ideal (the principal maximal ideal by assumption). It follows that $s^k=0$ for some $k\geq 1$ (the maximal ideal of an Artin local ring is nilpotent). The proper ideals of $R$ are then $(0),(s),\ldots,(s^{k-1}),R$, and any non-zero element of $R$ can be written as $us^j$ for $u\in R^\times$ a unit and a unique integer $j$ with $0\leq j\leq k-1$. Clearly such an element is a zero divisor if and only if $j\geq 1$.
To see why these facts hold (granting the fact that the maximal ideal of an Artin local ring is nilpotent, and assuming that $k\geq 1$ is the smallest integer such that $(s^k)=0$), let $I$ be a non-zero, proper ideal of $R$. Then $I\subseteq (s)$, and there is an integer $j$, $1\leq j\leq k-1$ such that $I\subseteq (s^j)$ but $I$ is not contained in $(s^{j+1})$ (since $I\neq 0$). Choose $r\in I$ such that $r\notin(s^{j+1})$. Then $r\in(s^j)$, so we can write $r=s^jr^\prime$ for some $r^\prime$. If $r^\prime\in(s)$ then $r\in (s^{j+1})$, so $r^\prime\notin(s)$, that is, $r^\prime$ is a unit. So $(s^j)=(r^\prime)\subseteq I$.
If any inclusion in the chain $R\supseteq(s)\supseteq(s^2)\supseteq\cdots(s^{k-1})\supseteq 0$ is not strict, i.e. if $(s^j)=(s^{j+1})$ for some $j$ with $0\leq j\leq k-1$, then $s(s^j)=(s^j)$ implies $(s^j)=0$ by Nakayama, a contradiction (since $R\neq 0$ and $k$ is the smallest integer $\geq 1$ with $(s^k)=0$).
Now if $r\in R\setminus\{0\}$, we have $r=(s^j)$ for a unique $j$ between $0$ and $k-1$. This means we can write $r=s^ju$ where $u\notin(s)$ (because $u\in(s)$ implies $r\in(s^{j+1})\subseteq(s^j)$, so $(s^j)=(s^{j+1})$, a contradiction as above), that is, $u\in R^\times$.
What's important in the above is not that $R$ is finite. I only used that to deduce that $R$ is Artinian. Everything I said is true for an arbitrary Artin local ring with principal maximal ideal.
I don't really understand your second question, because you say that $R$ has principal maximal ideal but then you list several generators.