Note that apparently James does no longer support this argument, although he never has acknowledged that he was wrong.

As before I will parafrase the proof to proof something that is false. Again, you can lay this proof side by side with James' proof at his page. I have comments interspeced in red.

Note, mentioning that x, y and z are coprime indicates that they are intended to be all integers; otherwise that makes no sense.

Here the original equation (x^p+y^p=n.z^p) is used. Square both sides and eliminate x^2p+y^2p.

The use Q and Q', here and later, is just a shorthand and the verbiage surrounding them is quite superfluous; I let it go by without comment.

This terminates the Q verbiage.
And upto this point everything is quite correct (although the formulation leaves something to be desired), also there is no proof for his assertion about Q', only an example, but a proof is not so very difficult.

Now this is not what Gauß' Fundamental Theorem of Algebra actually says. The proper formulation is:
Every polynomial of degree n with complex coefficients has n roots in the complex numbers.
In this case, because the equation is homogeneous in x, y and z, and because x and y always come together with the same power, we can see it as a polynomial in z²/xy, and so the FTA applies. Rewriting we can get what James states, except that where he writes a ring he should actually write the complex numbers, because that is the only thing that FTA guarantees.
In an article, Arturo Magidin actually proved that the numbers a_i and b_i can be chosen to be algebraic integers when the coefficients are integer (and his result is in fact a bit stronger). So the ring James could use would be the ring of algebraic integers, and I think that a smaller ring is not generally applicable.

Now I do not understand this at all. When we apply the substitutions in the last equation in the Q and Q' verbiage above we find on the right:
0 (mod 2)
On the left we find:
n²z^2p - 2
So we have:
n²z^2p - 2 = 0 (mod 2)
saying only that the left side is even.
This leaves something open: the modular equation only holds when x^p+y^p=z^p but this is false for integral z when x=1 and y=1. So this seems to imply that z is no longer integral, but in that case saying that the left side of the equation above is even can not apply.
Furthermore, the factors (a_iz² + b_ixy) can at best be algebraic integers. So saying their product is even does not say much.

Now something strange is happening, James suddenly shifts to v=-1, so the arguments above can not apply (the leading term of the polynomial in his original proof becomes 0, so it is a polynomial of smaller degree. He gets at a conclusion, and uses that to conclude something about arbitrary v. Note that although the formulas will not be the same with the integer factor n used above, the arguments are the same!

Note that this sharing holds if and only if x^p+y^p=n.z^p.
Note also that it is not clear at all that shifting names of x, y and z will indeed result in an F that is not equal to 1.

I admit that I do not understand this completely, nevertheless, this is false. Because the best we can say about the a_i and b_i, is that they are algebraic integers, and because there are algebraic integers that are units, there is nothing strange about a b being a divisor of an odd integer and of 2.

Therefore this is bogus.

It is also worth mentioning that this is only true whenever the condition x^p+y^p=z^p holds. So it must be a factor, not for all and every x, y and z, but only whenever x, y and z meet the condition. Moreover, whenever x, y and z do not meet the condition, the second expression can not be a factor of the first expression.