TITLE: $x^2+y^3 = z^4$ for positive integers
QUESTION [5 upvotes]: How can I solve this diophatine equation : 
$$x^2+y^3=z^4$$
for $(x, y, z) \in \mathbb{Z}_{>0}$
I tried to look on wolfram alpha yet it seems like there aren't any solutions...

REPLY [3 votes]: This is a special case of the generalized Fermat equation 
$$x^p+y^q=z^r$$
For $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}\le 1$ it has only finitely many coprime integer solutions, as has been proved by Darmon and Granville. For $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}>1$ however, there are infinitely many coprime nonzero integer solutions, given by a finite set of 2-parameter families, see Beukers, The Diophantine equation $Ax^p + By^q = Cz^r$, Duke Math. J. 91 (1998), 61-88. The explicit parameterizations (with proofs) can be found in Chapter 14 of Cohen's book
Number Theory, Vol. II.