= Solution
For one fixed <POVM> repeated on independently prepared copies, the <Born rule> gives $m$ outcome probabilities with one normalization constraint. An arbitrary mixed qubit <density operator> has three independent real parameters,
$$
\rho=\frac12(I+\mathbf r\cdot\boldsymbol\sigma),\qquad |\mathbf r|\leq1.
$$
The probability map is affine in $\mathbf r$. It can be injective on the full <Bloch ball> only if $m-1\geq3$. If its rank is smaller, a nonzero undetected direction gives two nearby physical Bloch vectors with the same probabilities. Hence $m\geq4$ for an <informationally complete POVM>.
To attain this bound, take the four directions
$$
\mathbf n_1=\frac{(1,1,1)}{\sqrt3},\quad
\mathbf n_2=\frac{(1,-1,-1)}{\sqrt3},\quad
\mathbf n_3=\frac{(-1,1,-1)}{\sqrt3},\quad
\mathbf n_4=\frac{(-1,-1,1)}{\sqrt3}.
$$
They are the vertices of a <regular tetrahedron>. The <tetrahedral qubit POVM> has effects $E_i=(I+\mathbf n_i\cdot\boldsymbol\sigma)/4$, each positive and rank one, and $\sum_iE_i=I$. The observed probabilities are
$$
p_i=\operatorname{Tr}(\rho E_i)=\frac{1+\mathbf r\cdot\mathbf n_i}{4}.
$$
Since $\sum_i\mathbf n_i=0$ and $\sum_i\mathbf n_i\mathbf n_i^T=4I_3/3$, they recover the state by
$$
\boxed{\mathbf r=3\sum_{i=1}^4p_i\mathbf n_i,\qquad m_{\min}=4.}
$$
This is the minimum for a single fixed measurement family. If the wording permits changing measurement settings between groups of copies, three binary projective measurements of the three Pauli observables also suffice, each with $m=2$. That procedure is <informationally complete three-observable qubit tomography>, and it does not contradict the four-outcome bound for one fixed POVM. Tomography estimates probabilities statistically from many copies; a finite sample does not determine an arbitrary continuous state exactly.
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