Solve $x^5 - 1 = y^2$ for integer values of $(x,y)$ .

So $$x^5=y^2+1=(y+i)(y-i).$$ Considering this modulo $4$ gives $y$ even and $x$ odd. The gcd of $y\pm i$ in the Gaussian integers divides $2$ and $y^2+1$ (which is odd) so it is $1$. As the Gaussian integers has unique factorisation and has four units, both $y\pm i$ are fifth powers, so $$y+i=(a+bi)^5=(a^5-10a^3b^2+5ab^4)+(5a^4b-10a^2b^3+b^5)i.$$ So $b\mid 1$ and $b=\pm1 $ and $$5a^4-10a^2+1=\pm1.$$ The only integer solution of this is $a=0$ leading to $y=0$ and $x=1$.

This is Mihailescu's theorem in the case $a=5, b=2$. The case $b=2$ was solved by Victor-Amédée Lebesgue in 1850.