$x^5 + y^2 = z^3$

Yes, there are infinitely many solutions. In fact, there are many parametrizations of the solutions.
According to a book${}^{\color{blue}{[1]}}$ on my bookshelf,

Up to changing $y$ into $-y$, there are exactly 27 distinct parametrizations of the equations $x^5 + y^2 = z^3$.

One of the simplest paremetrization is given by following formula.

$$\begin{align} x =&\; 12st(81s^{10}-1584t^5s^5-256t^{10})\\ y =&\; \pm (81s^{10} + 256t^{10})\\ &\;\;\times (6561s^{20} - 6088608t^5s^{15} - 207484416t^{10}s^{10} + 19243008t^{15}s^5 + 65536t^{20})\\ z =&\; 6561s^{20}+2659392t^{5}s^{15}+10243584t^{10}s^{10} - 8404992t^{15}s^5 + 65536t^{20} \end{align}$$

For example, following two random choices of $s,t$ give you two sets of relative prime solutions.

• $(s,t) = (1,1) \leadsto (x,y,z) = (-21108,-65464918703,4570081)$
• $(s,t) = (1,2) \leadsto (x,y,z) = (-7506024,127602747389962225,-196120763999)$

The book I have is actually quoting result from a thesis${}^{\color{blue}{[2]}}$ by J. Edwards. Consult that if you really want to get into the details.

References

• $\color{blue}{[1]}$ Henri Cohen, Number Theory Number Theory Volume II: Analytic and Modern Tools,
  $\S 14.5.2$ The Icosahedron Case $(2,3,5)$.

• $\color{blue}{[2]}$ J. Edwards, Platonic solids and solutions to $x^2+y^3 = dz^r$, Thesis, Univ. Utrecht (2005).

There is a beautiful connection between $a^5+b^3=c^2$ and the icosahedron. Consider the unscaled icosahedral equation,

$$\color{blue}{12^3u v(u^2 + 11 u v - v^2)^5}+(u^4 - 228 u^3 v + 494 u^2 v^2 + 228 u v^3 + v^4)^3 = (u^6 + 522 u^5 v - 10005 u^4 v^2 - 10005 u^2 v^4 - 522 u v^5 + v^6)^2\tag1$$

By scaling $u=12x^5$ and $v=12y^5$ (or various combinations thereof like $u=12^2x^5$, etc), we then get a relation of form,

$$12^5a^5+b^3=c^2$$