Here's the Rz matrix:
$$ Rz(\theta) = \begin{bmatrix} e^{-i\theta/2} & 0 \\ 0 & e^{i\theta/2} \end{bmatrix} $$
As I understand it, Rz rotates around the Z axis on the Bloch sphere. Since $|+\rangle$ and $|-\rangle$ are both on the Bloch sphere X-Y plane, it seems you should be able to rotate between them; however, I can't figure out a value of $\theta$ to do that.
Am I misunderstanding how the rotation gates work, or is there just a solution I'm not seeing?
The more elegant approach is to view $R_z(\phi)$ as a linear transformation acting on the basis $\{|0\rangle, |1\rangle\}$. $|0\rangle$ is mapped to $e^{i\phi/2}|0\rangle$ and $|1\rangle$ is mapped to $e^{-i\phi/2}|1\rangle$ by the transformation. As, $|+\rangle = \frac{1}{\sqrt 2}(|0\rangle +|1\rangle)$ it'd be mapped to $R_z(\phi)|+\rangle = \frac{1}{\sqrt 2}(e^{i\phi/2}|0\rangle + e^{-i\phi/2}|1\rangle)$. Now the question is, for what value of $\phi$ (if at all), $R_z|+\rangle$ coincides with $|-\rangle = \frac{1}{\sqrt 2}(|0\rangle - |1\rangle)$. As global phase factors are irrelevant in quantum mechanics, you just need to check for what value of $\phi$ (if at all) the coefficients of the basis vectors are in proportion i.e.,
$$\frac{e^{i\phi/2}}{e^{-i\phi/2}} = \frac{1}{-1} \implies e^{i\phi} = -1 \implies e^{i\phi} = e^{i\pi} \implies \phi = \pi$$
assuming $\phi \in [0, 2\pi)$.
Obviously, the Bloch sphere visualization (as in @kludg's answer) makes it even more evident. You can clearly see that a anti-clockwise rotation of $\phi = \pi$ (180 degrees) about the $z$-axis ($|0\rangle$ - $|1\rangle$ axis) takes the $|+\rangle$ state to the $|-\rangle$ state.
P.S: I used $R_z(\phi)$ instead of $R_z(\theta)$ in order to match the convention in the above diagram.
If you use $\theta = \pi$, you get the following:
$$ Rz(\pi)\begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{bmatrix} = \begin{bmatrix} e^{-i \pi/2} & 0 \\ 0 & e^{i \pi/2} \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{bmatrix} = \begin{bmatrix} -i & 0 \\ 0 & i \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{bmatrix} = -i\begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{-1}{\sqrt{2}} \end{bmatrix} $$
Which is equal to $|-\rangle$ under the global phase $-i = e^{-i\pi/2}$.
Just think where $|+\rangle$ and $|-\rangle$ states on the Bloch sphere ($x$ axis):
On IBM Q the Rz gate is defined as
\begin{equation} Rz(\theta) = \begin{pmatrix} 1 & 0 \\ 0 & \mathrm{e}^{i\theta} \end{pmatrix} \end{equation}
If you put $\theta = \pi$ then the matrix turns to
\begin{equation} Rz(\pi) = \begin{pmatrix} 1 & 0 \\ 0 & \mathrm{e}^{i\pi} \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} =Z \end{equation} Hence $Rz(\pi)|+\rangle = |-\rangle$.
You can rewrite your matrix as
\begin{equation} Rz(\theta) = \begin{pmatrix} \mathrm{e}^{-i\frac{\theta}{2}} & 0 \\ 0 & \mathrm{e}^{i\frac{\theta}{2}} \end{pmatrix} = \mathrm{e}^{-i\frac{\theta}{2}} \begin{pmatrix} 1 & 0 \\ 0 & \mathrm{e}^{i\theta} \end{pmatrix} \end{equation}
Hence the only difference is a global phase.
Overall, Rz gate matrix as defined on IBM Q seems to more convinient as you do not have to deal with a global phase coeficient.
