Well I am kinda curious
I am sure there is an easier method but I definitely can't find it
JeffM can you show us your method ? or give a hint ?
I suspect as well that there is an easier method, but I also thought that difference of powers would be the key, perhaps with assistance from the integer root theorem. Here basically is as far as I have got along this line.
It is obvious that x = 1 is not a solution because
$1^4 + 1^3 + 1^2 + 1 + 1 = 5,\ not\ a\ perfect\ square.$
So we can assume henceforward that $x \ne 1\ and\ \dfrac{x^5 - 1}{x - 1} = n.$
So far this is easy, but now things get hairy. Here is the route I tried, now excluding x equal 1, and
assuming 1 < n = m^2 And m > 1.
$\dfrac{x^5 - 1}{x - 1} = n = m^2 \implies x^5 - m^2x + (m^2 - 1) = 0.$
The quintic equation above is sure to have at least one real solution. So it may be factorable into a linear term and a quartic term.
One possible factoring is $(x - r)(x^4 + rx^3 + r^2x^2 + px - q) \implies qr = m^2 - 1.$
$r = 0 \implies m^2 - 1 = 0 \implies m = 1.$
That works $0^4 + 0^3 + 0^2 + 0 + 1 = 1 = 1^2.$ Going forward we can assume r and so x is not zero.
$q = \dfrac{m^2 - 1}{r}.$ If r is an integer, then q is rational.
$-\ m^2x = -\ prx - qx \implies m^2 = pr + q$ because x = 0 has already been addressed.
If r is an integer, p is rational.
$m^2r = pr^2 + qr = pr^2 + + m^2 - 1 \implies m^2r - m^2 = pr^2 - 1 \implies$
$pr^2 - m^2r + m^2 - 1 = 0 \implies r = \dfrac{m^2 \pm \sqrt{m^4 - 4p(m^2 - 1)}}{2p} = \dfrac{m^2 \pm \sqrt{m^4 - 4pm^2 + 4p}}{2p}.$
The only way to make $m^4 - 4pm^2 + 4p$ a perfect square is for p = 1.
In that case $r = \dfrac{m^2 \pm \sqrt{(m^2 - 2)^2}}{2} \implies r = m^2 + 1\ or\ r = 1.$
But that answer is impossible. So this factoring can be excluded.
I suspect that if you continue along this line you get the correct answer. It's Thanksgiving weekend and I have family to attend to. Be back later.