Picking up a simplified version of Jason Starr's idea:

$R = \mathbb{Z}_p \times \mathbb{Z}$ has dimension $1\,\,( \mathbb{Z}_p$ denotes the p-adic integes) and Jacobson radical $(p) \times 0 \not= 0 = Nil(R)$. Hence $R/J(R)=\mathbb{F}_p \times \mathbb{Z}$ has dimension $1$ as well.
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**Background:** The prime ideals of the product $R=R_1 \times R_2$ of commutative rings are $p_1 \times R_2$ and $R_1 \times p_2$. These are maximal iff $p_i$ are maximal in $R_i$. This shows:

- $\dim R = \max(\dim R_1, \dim R_2)$
- $J(R) = J(R_1) \times J(R_2)$
- $\dim R/J(R) = \max( \dim R_1/J(R_1), \dim R_2/J(R_2))$

In particular any domains $R_1, R_2$ with $J(R_1) \neq 0 = J(R_2)$ and $\dim R_2 \ge \dim R_1$ will give a counterexample.