Ignoring boundary conditions, the PDE has solutions $\phi = a\; \text{sn}\left(b x, c\right)$ where $\text{sn}$ is the Jacobi SN function (in Maple's parametrization: note that Mathematics uses a different convention), ${a}^{2}=2\;{\dfrac {{b}^{2}-{k}^{2}}{\lambda}}$, ${c}^{2}={\dfrac {{k}^ {2}}{{b}^{2}}}-1$. These are periodic in $x$, with a period depending on $b$ and $c$. For any given $k$ and $\lambda$ (at least in some region) I believe there should be infinitely many values of $a,b,c$ for which the period divides $2 \pi$. For example, with $k=2$ and $\lambda=1$ I get a period dividing $2 \pi$ for $c = 2.722857918$, $3.502129242$, $4.303773851$, $5.116503598$, etc. Of course we can replace $bx$ by $\sum_j b_j x^j$ with $\sum_j b_j^2 = b^2$.
EDIT: We can think about it this way. Assume $k, \lambda > 0$.
By scaling $y$ and time $t$ we can non-dimensionalize the autonomous differential equation $\ddot{y} + k^2 y + \lambda y^3 = 0$ to $\ddot{y} + y + y^3 = 0$.
The phase-plane trajectories of this autonomous differential equation are the closed curves $\dfrac{v^2}{2} + \dfrac{y^2}{2} + \dfrac{y^4}{4} = C$ for $C > 0$, where $v = dy/dt$. The period $P$ is of the orbit through $(y=y_0, v=0)$ is, by symmetry, $4$ times the time needed to get from $(y_0,0)$ to $(0,\sqrt{ y_0^2 + y_0^4/2})$, and thus
$$ P = 4 \int_0^{y_0} \dfrac{dy}{\sqrt{ y_0^2 - y^2 + y_0^4/2 - y^4/2}}$$
Under the change of variables $y = s y_0$ this becomes
$$ P = 4 \sqrt{2} \int_0^1 \dfrac{ds}{\sqrt{2 - 2 s^2 + y_0^2 - y_0^2 s^4}} $$
which goes to $0$ as $y_0 \to \infty$.
