$x^2+y^3 = z^4$ for positive integers 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...
 A: 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.
A: $$X^{n}+Y^{n+1}=Z^{n+2}$$
Solution always can be written, for example.
$$X=(c^2-b^2)^{(n+2)}b^{(n^2+2n-1)}c^{(n+1)^2}$$
$$Y=(c^2-b^2)^{(n+1)}b^{(n-1)(n+2)}c^{n(n+1)}$$
$$Z=(c^2-b^2)^{n}b^{(n-1)(n+1)}c^{(n^2+1)}$$
If you make this change.  $X^2+Y^3=Z^4$
$$p=tz(2zk^2+t)$$
$$s=tzk^2(2zk^2-t)$$
The result of such decision.
$$X=sp^3$$
$$Y=2tzk^2p^2$$
$$Z=kp^2$$
Where the number $t,z,k$ - integers and set us.  You may need after you get the numbers, divided by the common divisor.
