Borrowing from 4 different comments to make a complete answer (and making it community wiki):
We can construct a bijection between the set of del Pezzo surfaces and $\pi_4(S_3)$ using topology. However, the proof that this is a bijection depends on the classification of del Pezzo surfaces and does not give an explanation for why there are exactly two.
Both del Pezzo surfaces of degree $8$ are $S^2$-bundles over $S^2$, by the projection map $\mathbb P^1 \times \mathbb P^1 \to \mathbb P^1$, or by projection from the blown-up point in $\mathbb P^2$ onto a line.
$S^2$-bundles over a manifold $M$ are classified by $$Map(M, BDiff(S^2))= Map(M, BO(3))$$ Thus $S^2$ bundles over $S^2$ are classified by $$Map(S^2, BO(3))= \pi_2 (BO(3)) = \pi_1(O(3)) = \pi_1(SO(3)) = \mathbb Z/2$$
Furthermore the map from del Pezzo surfaces to $S^2$-bundles on $S^2$ is bijective, because the two surfaces are topologically distinct. We can check this by computing the intersection form. By the Kunneth formula $H^2(\mathbb P^1 \times \mathbb P^1)$ has generators $H$ and $V$ with $H \cdot H = 0$, $H \cdot V = 1$, $V \cdot V=0$. On the other-hand by the blow-up formula $H^2(Bl_x( \mathbb P^2))$ has generators $H$ and $E$ with $H \cdot H = 1$, $H \cdot E = 0$, $E \cdot E=-1$. The first intersection form is even, while the second is odd, so the two spaces are not homoeomorphic.
Finally $\pi_4(S^3)$ is the first stable homotopy group of spheres, which by the Pontrjagin construction is the same as the group of framed $1$-cobordisms $\Omega_1^{fr}$. $\pi_1( SO(3))$. A framing on a circle is just an element of $\pi_1(SO) = \pi_1(SO(3))= \mathbb Z/2$. In fact this gives an isomorphism between the stable homotopy group and $\pi_1(SO(3))$.
Composing these bijections, we get a bijection between degree 8 del Pezzo surfaces and $\pi_1(SO(3))$.