$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$ $\newcommand{\arcsinh}{\operatorname{arcsinh}}$
Consider problem \begin{align} & \Delta u-cu =f && \text{in }\mathcal{D}, \label{eq-7.1.1}\\[3pt] & u=g && \text{on }\ \Gamma_- \label{eq-7.1.2}\\[3pt] & \partial_\nu u -\alpha u =h && \text{on }\ \Gamma_+\label{eq-7.1.3} \end{align} where $\mathcal{D}$ is a connected bounded domain, $\Gamma$ its boundary (smooth), consisting of two non-intersecting parts $\Gamma_-$ and $\Gamma_+$, and $\nu$ a unit interior normal to $\Gamma$, $\partial_\nu u:=\nabla u\cdot \nu$ is a normal derivative of $u$, $c$ and $\alpha$ real valued functions.
\begin{align*} - &\int_\mathcal{D} fu\,dxdy = -\int_\mathcal{D} (u\Delta u-cu^2)\,dxdy\\[3pt] =&\int_\mathcal{D} (|\nabla u|^2+cu^2)\,dxdy +\int _\Gamma u\partial_\nu u\,ds\\[3pt] =&\int_\mathcal{D} (|\nabla u|^2+cu^2)\,dxdy + \int _{\Gamma_-} g\partial_\nu u\,ds + \int _{\Gamma_+} (\alpha u^2+ hu) \,ds \end{align*} where to pass from the first line to the second one we used equality $u\Delta u = \nabla (u\cdot u) -|\nabla u|^2$ and Gauss formula, and to pass to the third line we used $u=g$ on $\Gamma_-$ and $\partial_\nu u=\alpha u+h$ on $\Gamma_+$. Therefore assuming that \begin{equation} c\ge 0,\qquad \alpha \ge 0 \label{eq-7.1.4} \end{equation} we conclude that $f=g=h0\implies \nabla u=0$ and then $u=\const$ and unless \begin{equation} c\equiv 0,\quad \alpha\equiv 0, \qquad \Gamma_-=\emptyset \label{eq-7.1.5} \end{equation} we conclude that $u=0$.
So, if (\ref{eq-7.1.4}) is fulfilled but (\ref{eq-7.1.5}) fails problem (\ref{eq-7.1.1})--(\ref{eq-7.1.3}) has no more than one solution (explain why). One can prove that the solution exists (sorry, we do not have analytic tools for this).
Theorem 1. If (\ref{eq-7.1.4}) is fulfilled but (\ref{eq-7.1.5}) fails problem (\ref{eq-7.1.1})--(\ref{eq-7.1.3}) is uniquely solvable.
Assume now that (\ref{eq-7.1.5}) is fulfilled. Then $u=C$ is a solution with $f=h=0$. So, problem has no more than one solution modulo constant. Also \begin{equation*} \int_\mathcal{D}f\,dxdy= \int_\mathcal{D}\Delta u\,dxdy= -\int_\Gamma \partial_\nu u \,ds \end{equation*} and therefore solution of \begin{align} & \Delta u =f && \text{in }\ \mathcal{D},\label{eq-7.1.6}\\[3pt] & \partial_\nu u =h && \text{on }\ \Gamma\label{eq-7.1.7} \end{align} does not exist unless \begin{equation} \int_\mathcal{D}f\,dxdy+\int_\Gamma h \,ds=0. \label{eq-7.1.8} \end{equation} One can prove that under assumption (\ref{eq-7.1.8}) the solution exists (sorry, we do not have analytic tools for this).
Theorem 2.
Remark 1. Sure, these arguments and results hold in any dimension.