Let $(R,\mathfrak m)$ be a Noetherian local ring with at least one non-zero-divisors in $\mathfrak m$. Let $a\in \mathfrak m$ and $p\in \mathbb N$. Then we can find $a'\in \mathfrak m^p$ such that $a+a'$ is a non-zero-divisor.
By this question we know $\mathfrak m$ can be generated by nonzero divisors, hence so is any power of $\mathfrak m$. Let $k$ be such that $a\in \mathfrak m^k\setminus \mathfrak m^{k+1}$. Then we can write $a=\sum r_im_i$, $m_i\in \mathfrak m^k$ nonzerodivisors. Then since $a\notin \mathfrak m^{k+1}$, at least one $r_j$ must lie outside $\mathfrak m$, hence invertible. The $a-\sum_{i\neq j} r_im_i=r_jm_j$ is a non-zero-divisor. So this settles the question for $p\le k$. But can this idea be pushed further to show for $p> k$ or is there any other approach? Thanks.
Let $\mathfrak p_1,...\mathfrak p_s$ be all the associated primes and we can assume they are incomparable by discarding the ones contained in the others (since we only care about their union). We list them in such a way that $a\in \mathfrak p_1\cap\cdots\cap\mathfrak p_r$ and $a\notin \mathfrak p_{r+1}\cup\cdots\cup\mathfrak p_s$ for $0\le r\le s$. Then since $R$ has positive depth, we can pick $b\in \mathfrak m^p\setminus (\mathfrak p_1\cup\cdots\cup\mathfrak p_s)$. By prime avoidance, we can also pick $c\in (\mathfrak p_{r+1}\cap\cdots\cap \mathfrak p_s)\setminus (p_1\cup\cdots\cup\mathfrak p_r)$. Then $a':=bc\in \mathfrak m^p$ satisfies $a+a'\notin \mathfrak p_1\cup\cdots\cup\mathfrak p_s$ , so it is not a zero divisor.