jmc

By applying the difference of squares factorization repeatedly, we get $$\begin{array}{rl}{x}^8-1& =(x^4-1)(x^4+1)\\ & =(x^2-1)(x^2+1)(x^4+1)\\ & =(x-1)(x+1)(x^2+1)(x^4+1).\end{array}$$We can factor $x^4+1$ further with a clever application of difference-of-squares: $$\begin{array}{rl}{x}^4+1& =x^4+2x^2+1-2x^2\\ & =(x^2+1{)}^2-(x\sqrt{2}{)}^2\\ & =(x^2+x\sqrt{2}+1)(x^2-x\sqrt{2}+1).\end{array}$$Thus, $$x^8-1=(x-1)(x+1)(x^2+1)(x^2+x\sqrt{2}+1)(x^2-x\sqrt{2}+1).$$The quadratic factors have no real roots, so the factorization can have at most $\boxed{5}$ factors.