= Solution
Here a domain is connected. Suppose both a negative point and a zero point exist. The open set $D=\{v<0\}$ then has a boundary point $q$ inside $\Omega$. Choose $z\in D$ close enough to $q$ that the <Euclidean ball> of radius $R=\operatorname{dist}(z,Z)$ stays compactly inside $\Omega$; it is contained in $D$ and touches $Z$ at some $y$. Attainment of this distance follows by restricting to a compact neighbourhood of $q$.
On that <Euclidean ball> $v$ is twice differentiable and
$$
\Delta v=-v>0,\qquad v(y)=0>v(x)\quad(x\in B_R(z)).
$$
The <Hopf boundary point lemma> gives a strictly positive outward radial derivative at $y$. But $y$ is an interior maximum of the $C^1$ function $v$ on $\Omega$, so $Dv(y)=0$. This contradiction proves the <Hopf dichotomy with classical regularity away from the zero set>:
$$
\boxed{v\equiv0\quad\text{or}\quad v<0\text{ throughout }\Omega.}
$$
For the requested weaker-regularity counterexample take $\Omega=(-1,1)^n$ and
$$
\boxed{v(x)=-\max\{\sin x_1,0\}.}
$$
It is a <locally Lipschitz function> and nonpositive. Its negative set is $x_1>0$, where $v=-\sin x_1$ is smooth and $\Delta v+v=0$. On $x_1\leq0$ it is zero. The <normal derivative> jumps from zero to minus one across $x_1=0$, so it is not $C^1$. All other hypotheses hold; imposing the equation also across that interface would be an additional condition absent from the question.
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