In "Grobner Bases and the Computation of Group Cohomology" Hypothesis 1.5 is:
Let $k$ be a field of characteristic $p$. Let $\Lambda$ be a finite dimensional $k$-algebra (associative with one) satisfying $\Lambda/\text{rad}(\Lambda)\cong k$ [where $\text{rad}(\Lambda)$ is the Jacobson radical of $\Lambda$]. Suppose we are given a presentation $0\to I \to k\langle\mathbf{X}\rangle\overset{\phi}{\to}\Lambda\to 0$ of $\Lambda$ with the following properties: The set $\mathbf{X}$ is finite and the two-sided ideal $J\subseteq k\langle\mathbf{X}\rangle$ generated by $\mathbf{X}$ is the inverse image of $\text{rad}(\Lambda)$ [where $k\langle\mathbf{X}\rangle$ is the (non-commutative) $k$-algebra generated by $\mathbf{X}$].
Following this they make the following Remark 1.7:
Assuming Hypothesis 1.5, Nakayama's Lemma says there is an $n_I\geq 1$ satisfying $J^{n_I}\subseteq I\subseteq J$.
Since we know $J$ is a finitely generated maximal ideal, if we know $I\subseteq J$ then $J^{n_I}\subseteq I\subseteq J$ I thought follows from something like this question (although I'm not sure if some part of that fails in the noncommutative case).
If that's the case, how does Nakayama's Lemma imply $I\subseteq J$?