In Diophantine equations over the twentieth century: a (very) brief overview , p. 5

Problem Let $f(x) \in \mathbf{Z}[x]$ be an irreducible polynomial of degree at least 2. Do the Diophantine equation $$f(x)=y^2z^3$$ have only finitely many solutions in non-zero integers $x,y$ and $z$?

The most trivial case is $z=1,f(x)=ax^2+1$ where $a$ is not square.

This leads to the Pell equation $y^2-ax^2=1$, which has infinitely many solutions.

Another approach is let $f(x)=x^2+1$. For fixed $z$, this leads to Pell equation $x^2-z^3 y^2= -1$. For infinitely many $z$, it has infinitely many solutions $x,y$.

Couldn't find the reference online.

Is the problem misquoted?

In Diophantine equations over the twentieth century: a (very) brief overview , p. 5

Problem Let $f(x) \in \mathbf{Z}[x]$ be an irreducible polynomial of degree at least 2. Do the Diophantine equation $$f(x)=y^2z^3$$ have only finitely many solutions in non-zero integers $x,y$ and $z$?

The most trivial case is $z=1,f(x)=ax^2+1$ where $a$ is not square.

This leads to the Pell equation $y^2-ax^2=1$, which has infinitely many solutions.

Another approach is let $f(x)=x^2+1$. For fixed $z$, this leads to Pell equation $x^2-z^3 y^2= -1$. For infinitely many $z$, it has infinitely many solutions $x,y$.

Couldn't find the reference "[13] A. Schinzel and R. Tijdeman, On the equation $y^ m = f(x)$, Acta Arith. 31 (1976), 199-204." online.

Is the problem misquoted?