Why, in an applied magnetic field, is electron with spin parallel to the magnetic field higher?
In addition to the other answer, the energy  of a magnetic moment in a magnetic field is $E=-\bar{\bf{\mu}}.\bar{\bf{B}}$ where the dot indicates the dot product between the two vectors $\bar{\bf{\mu}}$ and $\bar{\bf{B}}$, thus the energy is lowest when the magnetic dipole is parallel to the field $B_0$; $E=-\mu B_0$. The sign of $\mu$ is opposite for nuclear and  electron spin angular momentum.The magnetic moment of a nuclear spin is defined as $\mu=+\gamma I$, where $I$ is the nuclear angular momentum and if the field is along only the $z$ direction  $\mu_z=\gamma I_z=m_z\gamma \hbar$ (Joule/Tesla) where $\gamma$ is the magnetogyric ratio, (unit radians/Tesla/sec) the components are $m_z$, the projection, magnetic or azimuthal spin quantum number on the $z$ axis, the axis parallel to the magnetic field $B_0$. $I_z$ is the $z$ component of the nuclear spin angular momentum. (The magnetogyric ratio is $\gamma = g\mu _0/\hbar$ where $\mu_0$ is the nuclear magnetron and $g$ the nuclear spin factor, different for each type of nucleus). The energy is then $E=-m_z\gamma\hbar B_0 $. For a proton $m_z=\pm 1/2$.

The spin magnetic moment of the electron is defined in a similar way to that for nuclear spin but with a negative sign; $\mu_e=-\gamma _e I_e$ where $I_e$ is the electron spin angular momentum and $\gamma _e$ the electron magnetogyric ratio. (  $\gamma _e=g_e\mu_B/\hbar$,  $g_e$ is the electron $g$ factor, $g_e =2.0023..$, and $\mu _B$ the Bohr magnetron. Some authors give a negative $g_e$, in this case the negative sign is removed from $\mu _e$).
(The nuclear magnetogyric ratio can be positive or negative, e.g. $\ce{26.75.10^7 rad T^{-1} s^{-1}}$ for protons. For a few types of nuclei e.g. $\ce{^{15}N, ^{29}Si}$ $\gamma$ is negative. The magnetogyric ratio for the electron is much $\ce{larger 1.76 10^11 rad T^{-1} s^{-1}}$.)