= Solution
In <spherical coordinates>, the angular <Laplacian> has <spherical harmonics> as <eigenfunctions>, $L^2Y_{lm}=\hbar^2l(l+1)Y_{lm}$ and $L_zY_{lm}=\hbar mY_{lm}$. Separation leaves
$$
-\frac{\hbar^2}{2m}\left[R''+\frac2rR'-\frac{l(l+1)}{r^2}R\right]-\frac{C}{r}R=ER,
\qquad C=\frac{e^2}{4\pi\epsilon_0}.
$$
Regularity and normalizability give <principal quantum number> $n=1,2,\ldots$, orbital <quantum number> $l=0,\ldots,n-1$, and <magnetic quantum number> $m=-l,\ldots,l$. Here the $m$ in the kinetic coefficient is the <electron mass>, distinct from the magnetic label.
For the proposed angular factor, $\cos\theta\propto Y_{10}$, so $l=1$ and <magnetic quantum number> zero. Direct differentiation gives $\nabla^2(re^{-\alpha r}\cos\theta)=(\alpha^2-4\alpha/r)\psi$. The $1/r$ term in the <Schrödinger equation> vanishes when $2\hbar^2\alpha/m=C$. With <Bohr radius> $a_0=\hbar^2/(mC)$,
$$
\boxed{\alpha=\frac1{2a_0},\qquad E=-\frac{\hbar^2\alpha^2}{2m}=-\frac{\hbar^2}{8ma_0^2}\simeq-3.40\,\mathrm{eV},\qquad l=1.}
$$
<wavefunction normalization> uses $\int d\Omega\cos^2\theta=4\pi/3$ and $\int_0^\infty r^4e^{-2\alpha r}dr=24/(2\alpha)^5$, giving
$$
\boxed{A=\sqrt{\alpha^5/\pi}=\frac1{4\sqrt{2\pi}\,a_0^{5/2}}.}
$$
An arbitrary constant phase is allowed. This is the $2p_z$ state. Under spatial inversion, $r$ is unchanged and $\cos\theta$ changes sign, so its <parity> <eigenvalue> is \b[$-1$], also given by $(-1)^l$.
Rotational invariance <means> rotated states have the same <energy>; it does not require each member of a degenerate <energy> eigenspace to be invariant. <rotations> mix the three $l=1$ magnetic substates. An orbital oriented along $z$ can therefore have an anisotropic <probability density> while the <Hamiltonian> remains spherically symmetric.
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