Prove that $x \in {\rm Jac}(R)$ if and only if $1-xy \in R^{\times}$ for all $y$.

Take $x \in Jac(R)$ and $y \in R$. Then see that $1-yx$ can't belong to any maximal ideal so the ideal generated by $1-yx$ is the whole ring $R$. You should be able to conclude that $1-yx \in R^\times$

Suppose that $x\in\text{Jac}(R)$, but there exists $y\in R$ such that $1-xy$ is not invertible. Then, there exists a maximal ideal $\mathfrak{m}$ such that $1-xy\in\mathfrak{m}$. Set $m=1-xy$. By hypothesis, $x\in\text{Jac}(R)\subseteq\mathfrak{m}$, so $x\in\mathfrak{m}$. But $\mathfrak{m}$ being an ideal implies that $xy\in\mathfrak{m}$, and then $$ m+xy=1-xy+xy=1\in\mathfrak{m}, $$ which is absurd because $\mathfrak{m}$ is a proper ideal.