How many ordered triples $(x,y,z)$ of real numbers are there such that $x+y^2=z^3$, $x^2+y^3=z^4$, and $x^3+y^4=z^5$ How many ordered triples $(x,y,z)$ of real numbers are there such that 
$x+y^2=z^3$
$x^2+y^3=z^4$
$x^3+y^4=z^5$?
Adding those equations together and factorizing gives you $x(1+x+x^2)+y^2(1+y+y^2)=z^3(1+z+z^2)$
Which is of the same form as the 1st equation, but I don’t know how to use this to my advantage. Hints, suggestions and solutions would be appreciated.
Taken from the 2018 BIMC

Hint
From $z^3\cdot z^5=(z^4)^2$ we obtain $$(x+y^2)(x^3+y^4)=(x^2+y^3)^2$$which leads to $$xy^2(x-y)^2=0$$
