The 1-qubit Bell states are given by
\begin{align*}
|{+}\rangle &= \frac{1}{\sqrt{2}} \left(|{0}\rangle + |{1}\rangle \right) & |{-}\rangle = \frac{1}{\sqrt{2}} \left(|{0}\rangle - |{1}\rangle \right)
\end{align*}
These are ortho-normal eigen-states of \(X\):
\begin{align*}
X |{+} \rangle &= +1 |{+} \rangle & X |{-} \rangle &= -1 |{-} \rangle
\end{align*}
Taking tensor products gives a 2-qubit ortho-normal basis:
\begin{align*}
|{+}{+} \rangle &\equiv |{+} \rangle \otimes |{+} \rangle = \frac{1}{2} \left(|{0}{0} \rangle + |{0}{1} \rangle + |{1}{0} \rangle + |{1}{1} \rangle \right) \\
|{+}{-} \rangle &\equiv |{+} \rangle \otimes |{-} \rangle = \frac{1}{2} \left(|{0}{0} \rangle - |{0}{1} \rangle + |{1}{0} \rangle - |{1}{1} \rangle \right) \\
|{-}{+} \rangle &\equiv |{-} \rangle \otimes |{+} \rangle = \frac{1}{2} \left(|{0}{0} \rangle + |{0}{1} \rangle - |{1}{0} \rangle - |{1}{1} \rangle \right) \\
|{-}{-} \rangle &\equiv |{-} \rangle \otimes |{-} \rangle = \frac{1}{2} \left(|{0}{0} \rangle - |{0}{1} \rangle - |{1}{0} \rangle + |{1}{1} \rangle \right)
\end{align*}
Note that, by construction, the four vectors in this 2-qubit basis do not describe 2-body entanglement since they are all tensor products. Linear combinations do, giving the 2-qubit Bell states:
\begin{align*}
|\Phi^{+} \rangle &= \frac{1}{\sqrt{2}} \left(|{0}{0}\rangle + |{1}{1}\rangle \right) = \frac{1}{\sqrt{2}} \left(|{+}{+}\rangle + |{-}{-}\rangle \right) \\
|\Phi^{-} \rangle &= \frac{1}{\sqrt{2}} \left(|{0}{0}\rangle - |{1}{1}\rangle \right) = \frac{1}{\sqrt{2}} \left(|{-}{+}\rangle + |{+}{-}\rangle \right) \\
|\Psi^{+} \rangle &= \frac{1}{\sqrt{2}} \left(|{0}{1}\rangle + |{1}{0}\rangle \right) = \frac{1}{\sqrt{2}} \left(|{+}{+}\rangle - |{-}{-}\rangle \right) \\
|\Psi^{-} \rangle &= \frac{1}{\sqrt{2}} \left(|{0}{1}\rangle - |{1}{0}\rangle \right) = \frac{1}{\sqrt{2}} \left(|{-}{+}\rangle - |{+}{-}\rangle \right)
\end{align*}
These states cannot be factorized as tensor products.
Another 1-qubit basis is the two eigen-states of \(Y\):
\begin{align*}
|{\circ}\rangle &= \frac{1}{\sqrt{2}} \left(|{0}\rangle + i|{1}\rangle\right) & |{\star}\rangle &= \frac{1}{\sqrt{2}} \left(|{0}\rangle - i|{1}\rangle\right)
\end{align*}
Taking tensor products gives another 2-qubit ortho-normal basis:
\begin{align*}
|{\circ}{\circ} \rangle & \equiv |{\circ} \rangle \otimes |{\circ} \rangle = \frac{1}{2} \left(|{0}{0} \rangle + i|{0}{1} \rangle + i|{1}{0} \rangle - |{1}{1} \rangle \right) \\
|{\circ}{\star} \rangle & \equiv |{\circ} \rangle \otimes |{\star} \rangle = \frac{1}{2} \left(|{0}{0} \rangle - i|{0}{1} \rangle + i|{1}{0} \rangle + |{1}{1} \rangle \right) \\
|{\star}{\circ} \rangle & \equiv |{\star} \rangle \otimes |{\circ} \rangle = \frac{1}{2} \left(|{0}{0} \rangle + i|{0}{1} \rangle - i|{1}{0} \rangle + |{1}{1} \rangle \right) \\
|{\star}{\star} \rangle & \equiv |{\star} \rangle \otimes |{\star} \rangle = \frac{1}{2} \left(|{0}{0} \rangle - i|{0}{1} \rangle - i|{1}{0} \rangle - |{1}{1} \rangle \right)
\end{align*}
I kind of like this notation: the star suggests complex conjugation.