First of all you have to get clear about the concepts. The (pure) states of a quantum system are described by a "state vector" (more precisely by a state vector modulo an arbitrary non-zero factor, a socalled ray, but that are finer details). In the socalled Schrödinger picture of time evolution the state vector ##|\psi(t) \rangle## is a function of time.
Observables are described by self-adjoint operators, and there is a particular operator, the Hamilton operator (usually representing the total energy of the system) which determines the time evolution of the state by the equations
$$\mathrm{i} \partial_t |\psi(t) \rangle=\hat{H} |\psi(t) \rangle.$$
If ##\hat{H}## is not explicitly time dependent a formal solution is
$$|\psi(t) \rangle=\exp \left (-\frac{\mathrm{i}}{\hbar} \hat{H} t \right ) |\psi(0) \rangle.$$
The values an observable ##A## can take when precisely measured are the eigenvalues of the corresponding self-adjoint operator ##\hat{A}##. Assuming that for any eigenvalue ##a## there is only one eigenvector (modulo a non-zero factor of course), ##|a \rangle## then the probability to get the value ##a## when measuring the observable at time ##t##, given the state ##|\psi(t) \rangle## is
$$P(t,a)=|\langle a|\psi(t)\rangle|^2.$$
The factors for both vectors have to be chosen such that they are normalized, i.e., ##\langle \psi|\psi \rangle=\langle a|a\rangle=1##.
Usually the eigenvectors of a self-adjoint operator form a complete orthonormal set (or a basis), i.e., you can expand any vector wrt. this basis of eigenvectors.
$$|\psi(t) \rangle=\sum_a |a \rangle \langle a|\psi(t) \rangle.$$
Now to solve the equation of motion given above (which is the Schrödinger equation in Dirac's bra-ket notation) a particularly convenient basis is the energy eigenbasis,
$$\hat{H} |E \rangle=E |E \rangle.$$
So we write
$$|\psi(t) \rangle=\sum_E |E \rangle \langle E|\psi(t) \rangle.$$
Now we have
$$|\psi(t) \rangle = \exp \left (-\frac{\mathrm{i}}{\hbar} \hat{H} t \right) |\psi(0) \rangle.$$
Plugging our decomposition in terms of energy eigenvectors in we simply get
$$|\psi(t) \rangle = \sum_E \exp \left (-\frac{\mathrm{i}}{\hbar} E t \right) |E \rangle \langle E |\psi(0) \rangle.$$
Particularly, if at time ##t=0## the state vector is an eigenvector (i.e., the system is prepared in an energy eigenstate) ##|\psi(0) \rangle=|E_0 \rangle## the equation simplifies to
$$|\psi(t) \rangle = \exp \left (-\frac{\mathrm{i}}{\hbar} E_0 t \right) |E_0 \rangle.$$
All that happens to the initial state vector is that it gets multiplied by a phase factor, but this doesn't really change the state, because all statevectors that only differ by just some non-zero factor describe the same (pure) state. This shows that the energy eigen states are the stationary states of the system. This distinguishes the energy eigenstates from any other basis.
Of course, often you have to describe the physics in terms of other bases. E.g., if you learn QT in terms of "wave mechanics" you start working in the position basis. Since the position operator ##\hat{x}## has continuous eigenvalues ##x \in \mathbb{R}## instead of sums you have to write integrals. Thus in this case the state vector is written as
$$\psi(t) \rangle=\int_{\mathbb{R}} \mathrm{d} x |x \rangle \langle x|\psi(t) \rangle,$$
and
$$\psi(t,x)=\langle x|\psi(t) \rangle$$
is the wave function. Now the Schrödinger equation is expressed for the wave function as
$$\mathrm{i} \hbar \partial_t \psi(t,x)=\hat{H} \psi(t,x),$$
where now ##\hat{H}## acts on the wave function. E.g., for a particle moving in some potential the Hamiltonian is indeed given as the total energy operator,
$$\hat{H}=\frac{1}{2m} \hat{p}^2+V(x)=-\frac{\hbar^2}{2m} \partial_x^2 + V(x).$$
From this you can evaluate the energy eigenvalues and eigenvectors by solving the eigenvalue equation (also known as the "time-independent Schrödinger equation")
$$\hat{H} u_E(x)=E u_E(x).$$
In Dirac notation
$$u_E(x)=\langle x|E \rangle.$$
The formal solution of the time-dependent Schrödinger equation then follows immediately:
$$\psi(t,x)=\langle x|\psi(t) \rangle=\sum_E \exp \left (-\frac{\mathrm{i}}{\hbar} t E \right ) \langle x|E \rangle \langle E|\psi(0) \rangle =\sum_E \exp \left (-\frac{\mathrm{i}}{\hbar} t E \right ) u_E(x) \psi_{E0},$$
where
$$\psi_{E0}=\langle E|\psi(0) \rangle = \int_{\mathbb{R}} \mathrm{d} x \langle E|x \rangle \langle x|\psi(0) \rangle = \int_{\mathbb{R}} \mathrm{d} x u_E^*(x) \psi_0(x).$$
