A reduced ring has no nonzero nilpotents.
In a commutative ring, any linear combination of nilpotent elements is nilpotent. Use the multinomial theorem and make n bigger than the sum of all the exponents that drive the individual nilpotents to 0.
If x is nilpotent, 1-x is a unit. This is because synthetic division terminates. That is, 1-xn is divisible by 1-x.
An element x is idempotent if x2 = x. If R is free of zero divisors, write x2-x = 0, whence x = 0 or 1.
Note that an idempotent x has an orthogonal counterpart 1-x. Each squared is itself, their sum is 1, and their product is 0. This works even in characteristic 2.
Conversely, let a finite set of orthogonal idempotents sum to 1, such that these idempotents commute with R. Let Ri be the principal ideal generated by ei. Show that each Ri is a subring, with ei acting as 1, and that R is isomorphic to the direct product of these subrings. The map is accomplished via x*1, where 1 is replaced with the sum of idempotents. Then, x*1 times y*1 expands into a large cross product, but all the mixed terms drop out, and we simply find e1*xy + e2*xy + … + en*xy, which is the projection of the product onto the component rings. All the algebra works; I'll leave it to you.
The algebra doesn't work when the orthogonal idempotents don't commute with R. Let a and b be idempotents in a ring with characteristic 2. Include x and y, such that xa = x, yb = y, ay = y, bx = x, and the other nonidempotent products are all 0. Verify this is a ring, with a+b = 1, that does not split into a finite product of subrings. Again, I'll leave the algebra to you.