Here is one for any $p>0$. Doing the usual change of unknown $v(x)=u(x)\exp(-\gamma(x)/2)$, you obtain
$$
-\Delta v + \left(\exp\left(-\frac{\gamma}{2}\right) \Delta \left(\exp\left(\frac{\gamma}{2}\right)\right)\right) v = v^p \exp\left(\frac{(p-1)\gamma}{2}\right)
$$
Note that
$$
\exp\left(-\frac{\gamma}{2}\right) \Delta \left(\exp\left(\frac{\gamma}{2}\right)\right)= \frac{1}{2}\Delta \gamma + \frac{1}{4}\left|\nabla\gamma\right|^2
$$
Let $\lambda(\gamma)$ be the first dirichlet eigenvalue of
$$
-\Delta \psi + \left(\frac{1}{2}\Delta \gamma + \frac{1}{4}\left|\nabla\gamma\right|^2\right) \psi = \lambda(\gamma)\psi
$$
It always exists if $\gamma \in C^2(\bar\Omega)$, but it could be positive or negative. However the corresponding eigenvector $\psi\in H^1_0(\Omega)$ can be taken postive.

Integrate by parts against $\psi$ to obtain
$$
\int_\Omega \left(\lambda(\gamma)v - \exp\left(\frac{(p-1)\gamma}{2}\right)v^{p}\right) \psi =0.
$$
If $\lambda(\gamma)<0$ the map
$$
z\to \lambda(\gamma)z -\exp\left(\frac{(p-1)\gamma}{2}\right) z^{p}
$$
is negative over $\mathbb{R}^{+}$ , and therefore a non-trivial $v$ leads to a contradiction, for any $p$. So one structural condition is

if $\lambda(\gamma)\leq 0$, then there is no nontrivial solution, for any $p$.

An explicit corollary is :

Let $\lambda_{+}(\Omega)$ be the first dirichlet eigenvalue of the Laplacian on $\Omega$. If
$$
\frac{1}{2}\Delta \gamma + \frac{1}{4}\left|\nabla \gamma\right|^2 \leq -\lambda_{+}(\Omega)
$$
there is no non-trivial positive solutions.

In the special case $p=1$, there could be non trivial solutions, of course, depending on $\Omega$. If $\lambda_{+}(\Omega)>1$, which happens for example when $\Omega\subset(0,\sqrt{d}\pi)^{d}$, choose $\gamma=2\ln(\psi)$ where $\psi$ is the solution of
$$-\Delta \psi = \lambda_{+}(\Omega^\prime) \psi \mbox{ on } \Omega^\prime,\quad \psi=0 \mbox{ on } \partial\Omega^\prime, \int_\Omega \psi=1
$$ for some $\Omega^\prime$ containing $\Omega$. The problem then become $$ -\Delta v = (\lambda_{+}(\Omega^\prime) +1)v \mbox{ in } \Omega.$$
As $\Omega^\prime$ grows, $(\lambda_{+}(\Omega^\prime) +1)$ will reach $\lambda_{+}(\Omega)$ at some point, leading to a non-trivial solution.
In general, if

$$
\frac{1}{2}\Delta \gamma + \frac{1}{4}\left|\nabla \gamma\right|^2 = \lambda_{+}(\Omega)-1,
$$
there is a non trivial solution for $p=1$.