Consider the basis
$$
B=\{\phi_{00},\ \theta_{0n},\ \xi_{n0},\ \phi_{mn},\ \psi_{mn},\ \theta_{mn},\xi_{mn}\}_{(m,n) \in \mathbb{N}^2}
$$ of $L^2([0,1]^2)$ given by
\begin{eqnarray}
\phi_{00}(x,y)&=&1,\
\theta_{0n}(x,y)=\sin(2\pi ny),\ \xi_{n0}(x)=\sin(2\pi nx),\\
\phi_{mn}(x,y)&=&\cos(2\pi mx)\cos(2\pi ny),\ \psi_{mn}(x,y)=\sin(2\pi mx)\sin(2\pi ny),\\
\theta_{mn}(x,y)&=&\cos(2\pi mx)\sin(2\pi ny),\ \xi_{mn}(x,y)=\sin(2\pi mx)\cos(2\pi ny).
\end{eqnarray}
Then for every $m,n \ge 0$ we have
$$
-\Delta\phi_{mn}=\lambda_{mn}\phi_{mn},\
-\Delta\psi_{mn}=\lambda_{mn}\psi_{mn},\
-\Delta\theta_{mn}=\lambda_{mn}\theta_{mn},\
-\Delta\xi_{mn}=\lambda_{mn}\xi_{mn}\ \forall\ m,n \ge 0,
$$
where
$$
\lambda_{mn}=4\pi^2(m^2+n^2) \quad \forall\ m, n\ge 0.
$$
This shows that the number $\lambda_{mn}$ is an eigenvalue of $-\Delta$ with corresponding eigenfunctions $\phi_{mn},\psi_{mn},\theta_{mn},\xi_{mn}$.
Because of the boundary conditions
\begin{eqnarray}
u(x,0)=u(x,1)&=&0 \quad \forall x \in [0,1],\\
u(0,y)=u(1,y)&=&0 \quad \forall y \in [0,1],
\end{eqnarray}
it is clear that a pair $(u,\lambda)$ satisfies the given PDE precisely when it belongs to $\{(c\psi_{mn},\lambda_{mn}): \ m,n \ge 1, \ c \in \mathbb{R}\setminus\{0\}\}$.
