 Solution

Let $D=\mathrm{dom}(f)$ and take a formula $\phi (x,\bar{d})$ over $D$. Since $T$ is a strongly minimal theory, the set $\phi (\mathcal{U},\bar{d})$ is finite or cofinite. If $\phi (b,\bar{d})$ holds, it cannot be finite because $b\notin \mathrm{acl}(D)$, so it is cofinite. Elementarity of $f$ makes $\phi (\mathcal{U},f\bar{d})$ cofinite. Its finite complement is algebraic over $f(D)$, so $c\notin \mathrm{acl}(\mathrm{ran}f)$ lies in the set. Applying the same argument to $\neg \phi$ gives the reverse implication. Hence $f\cup \{(b,c)\}$ is elementary: both elements realize the generic type in a strongly minimal theory over the corresponding domains.

If models $M,N$ have bases $B,C$ of the same dimension of a pregeometry, choose a bijection $B\to C$. Repeated use of the one-point result makes it elementary on every finite subset and hence on $B$. Every model is the algebraic closure of a basis. Extending the map back and forth across algebraic elements produces an isomorphism $M\to N$. Thus models of a strongly minimal theory with the same dimension are isomorphic.