Note that 3 → 4 → 1 is straightforward.
To show 1 → 2, select a minimal set of generators for a maximal ideal M. Let x be one of these generators, and write x as a product of irreducible elements. Since M is a prime ideal it contains at least one of these factors. Replace x with the prime factor that belongs to M. Do this for every generator in the set.
In a ufd, irreducible and prime are synonymous, and prime elements generate prime ideals. If M includes the generators x and y, then the prime ideals generated by x and y belong to M. However, there are no prime ideals properly contained in M, so x and y both generate M. We don't need both of them. If the set of generators is minimal, it consists of one generator, and M is principal. That completes 1 → 2.
Prove 2 → 3, and the first 4 criteria become equivalent. Assume the maximal ideals are principal. Each fractional ideal is a product of maximal ideals and their inverses, and the inverse of a principal ideal is principal. Thus all fractional ideals are principal, and 1 through 4 are equivalent.
Now 4 ⇔ 5 is a little off the beaten path, and if you're not familiar with torsion free modules you can skip to the next section. Let's take the forward direction first.
A finitely generated module over a pid is well characterized. In particular, a torsion free module becomes free, and that completes 4 → 5.
For the converse, assume finitely generated torsion free modules are free. Let H be any ideal in R. Now H is finitely generated and torsion free, hence it is free. Suppose H has rank > 1. A free R module of rank 2 now embeds in R, which is a free R module of rank 1. Let b and c act as a basis for a free R module of rank 2, inside H, inside R.
Let F be the fraction field of R. Let b and c generate an F module inside F. If a linear combination of b and c yields 0, multiply through by the common denominator, and b and c are not linearly independent over R. This is a contradiction, hence b and c are independent, and span an F vector space of dimension 2. This lives inside F, which is an F vector space of dimension 1, which is impossible. Therefore H has rank 1. It is a principal ideal, and R is a pid.