I asked Mathematica about the boundary behavior. First, the $P^{1/4}_{\nu } $ solution: We have, for $\epsilon \searrow 0$,
$$
(\sin \epsilon )^{1/4} P^{1/4}_{\nu } (\cos \epsilon ) = \frac{2^{1/4} }{\Gamma(3/4)} + O(\epsilon^{2} )
$$
and
\begin{eqnarray*}
(\sin (\pi -\epsilon ))^{1/4} P^{1/4}_{\nu } (\cos (\pi -\epsilon) ) &=& \frac{2^{3/4} \pi }{\Gamma(3/4)\Gamma(-\nu )\Gamma(1+\nu )} \\
& & -\frac{2^{1/4} \pi }{\Gamma(5/4)\Gamma(-1/4-\nu )\Gamma(3/4+\nu )} \sqrt{\epsilon } \\
& & + O(\epsilon^{2} )
\end{eqnarray*}
So, the $P^{1/4}_{\nu } $ solution automatically satisfies the boundary condition at $\theta =0$, whereas at $\theta =\pi $, we have to eliminate the term proportional to $\sqrt{\epsilon } $. That determines the eigenvalues: We need either $-1/4-\nu $ to be a negative integer or 0, or $3/4+\nu $ to be a negative integer or 0.
The specification of $\nu $ in the OP suggests the constraint $\nu \geq -1/2$; this excludes the second alternative, and therefore we obtain the spectrum $\nu = n-1/4$, $n=0,1,2,3,\ldots $.

The $Q^{1/4}_{\nu } $ solution, on the other hand, exhibits behavior proportional to $\sqrt{\epsilon } $ at $\theta=0$, with coefficient
$$
\frac{\pi^{2} }{2^{1/4} } \frac{\cos ((4\nu +1)\pi /8) \Gamma (-\nu /2 -1/8) \Gamma (\nu /2 +9/8) - \sin ((4\nu +1)\pi /8)\Gamma (-\nu /2 +3/8) \Gamma (\nu /2 +5/8)}{\Gamma (5/4) \Gamma (-\nu /2 -1/8) \Gamma (-\nu /2 +3/8) \Gamma (\nu /2 +3/8)\Gamma (\nu /2 +7/8)}
$$
A plot as a function of $\nu $ suggests that, for $\nu \geq -1/2$, this is positive and monotonically rising (I have not attempted to verify this analytically); the behavior proportional to $\sqrt{\epsilon } $ at $\theta=0$ can therefore not be eliminated, nor can it be compensated by admixture of the $P^{1/4}_{\nu } $ solution. Thus, there are no further solutions involving $Q^{1/4}_{\nu } $.

In summary, the complete spectrum is given by $\nu = n-1/4$, $n=0,1,2,3,\ldots $, or, in terms of $\lambda $, $\lambda = n(n+1/2)$, $n=0,1,2,3,\ldots $.