Answered shortly. It has to do with compactness. Boundary conditions are very relevant for instanton-like solutions since by definition they must have a finite action. This immediately implies that such solutions must have a particular decaying behavior. In this process what is done mathematically is, compactify space-time. However this procedure makes sense only if your norm is euclidean because the quadratic form $$x_\mu x^\mu = x_1^2 + x_2^2 + x_3^2 + x_4^2$$ is positive definite (technically $$SO(4)$$ symmetric which is a compact group), so a single point labeled as infinity can be defined, it representing all coordinates increasing and getting farther and farther from the origin. And then a solution can be found since expressions such as $$|x|^{-n}$$ make sense at infinity. This is not the case in the Minkowski case since its symmetry group is $$SO(3,1)$$ is not compact one cannot write solutions that fall-off consistently and give out finite actions.
Note: The above statements are not sufficient (just necessary) to have instanton solutions. Even in Euclidean signature in $$\mathbb{R}^4$$ with $$U(1)$$ symmetry, one does not have not trivial solutions because of gauge-equivalence.
• Thanks a lot for your answer. While this argument for the existence of BPST instanton in Euclidean space is standard (a map from the boundary at infinity to the field space for pure gauge field configurations), it is still not clear to me why the symmetry group $SO(3,1)$ does not allow a solution in Minkowski space? – Wein Eld Apr 23 '20 at 10:14