It has not so much to do with the Schrödinger equation, but with the eigenvalue problem for the Hamiltonian \hat{H}. In position space the Hamilton operator for a free particle is given by the differential operator
\hat{H}=-\frac{\hbar^2}{2m} \Delta.
It's indeed easy to schow that the plane wave
u_{\vec{k}}(\vec{x})=N \exp(\mathrm{i} \vec{k} \cdot \vec{x})
is a generalized eigenfunction of \hat{H} (with N an arbitrary constant). Just take the derivatives and check that it fulfills the eigenvalue equation
\hat{H} u_{\vec{k}}(\vec{x})=E_{\vec{k}} u_{\vec{k}}(\vec{x}).
You'll easily find the energy eigenvalue.
Also think about, whether this function can ever represent a state of the particle in the sense of quantum theory. To help a bit: The answer is a clear no!
Sometimes the eigenvalue equation for the Hamilton operator is called "the time-independent Schrödinger equation". Indeed, the relation with the Schrödinger equation,
\mathrm{i} \hbar \partial_t \psi(t,\vec{x})=\hat{H} \psi(t,\vec{x}),
is that the function
\psi_{\vec{k}}(t,\vec{x})=u_{\vec{k}}(\vec{x}) \exp \left (-\mathrm{i} \frac{t E_{\vec{k}}}{\hbar} \right)
is a solution. The eigenfunctions of the Hamilton operator represent the "stationary solutions", because it is constant in time up to the phase factor \exp(-\mathrm{i} t E_{\vec{k}}/\hbar).
