Let $I$ be an ideal. $I$ is called dense if its annihilator is $0$.
Given a dense ideal $I = a + J$, also $a^n + J$ is dense for all $n$. Because if $b(a^n + J) = 0$ then $ba^{n-1}I = 0$ so that $ba^{n-1} = 0$, and repeating we get $b = 0$. It follows that if $I = (a_1, \ldots, a_k)$ is dense, then $I = (a_1^{n_1}, \ldots, a_n^{n_k})$ is dense for any $(n_1, \ldots, n_k) \in \mathbb{N}^k$.
Does the general case hold when $I$ is not assumed finitely generated?
A counterexample would have to come from a ring with nilpotents.
In trying to prove it in the non finitely generated case, I was trying to look at the set of functions $f: \mathcal{I} \rightarrow \mathbb{N}$ such that the corresponding ideal $(a_1^{f(1)}, a_2^{f(2)}, \ldots)$ is not dense, where $\mathcal{I}$ indexes the generators of the dense ideal $I = (a_1, a_2, \ldots)$. We can partially order this set by domination. Chains of elements $f_1 \geq f_2 \geq \cdots$ have a natural candidate for a lower bound given by the function defined coordinatewise as $f(i) = \min(f_\alpha(i))$ where $f_\alpha$ ranges over the chain. If this lower bound was also in the set, then by Zorn's lemma I'd arrive at a contraction and the proof would be done. But that's a big if.