Solution: Let $R$ be a commutative ring and \[ \mathcal{N} \coloneqq \left\{ r \in R : \exists~ n \in \mathbb{N} \text{ such that } r^n = 0 \right\}. \] We claim that $\mathcal{N} $ is a subring of $R$. Let $r_1, r_2\in \mathcal{N} $. This means there exist positive integers $n_1$ and $n_2$ such that $r_1^{n_1} = 0 = r_2^{n_2}$. Note that if $r^k = 0$, then $(-r)^k = 0$, so $\mathcal{N} $ is closed under additive inverse. Consider \begin{align*} \left( r_1 - r_2 \right) ^{n_1 + n_2} & = \sum_{i=0}^{n_1 + n_2} \binom{n_1 + n_2}{i} r_1^{n_1 + n_2 - i} (-r_2)^i \\ & = \sum_{i=0}^{n_2} \binom{n_1 + n_2}{i} r_1^{n_1 + n_2 - i} (-r_2)^{i} + \sum_{i=n_2}^{n_1 + n_2} r_1^{n_1 + n_2 - i} (-r_2)^i. \end{align*} Since for any $0 \leq i \leq n_2$ we have $n_1 \lt n_1 + n_2 - i \lt n_1 + n_2$, and therefore, $r_1^{n_1 + n_2 - i} = 0$. Similarly, $(-r_2)^{i} = 0$. Thus, $(r_1 - r_2)^{n_1 + n_2} = 0$. Hence, $r_1 - r_2 \in \mathcal{N} $.

Now we will show that the product of $r_1$ and $r_2$ is also nilpotent. Recall that for a commutative ring $(ab)^k = a^k b^k$. So we have \begin{align*} (r_1 r_2)^{n_1} = r_1^{n_1} r_2^{n_2} = 0. \end{align*} Thus, $\mathcal{N} $ is closed under multiplication and hence $\mathcal{N} $ is a subring of $R$.