Does the Diophantine equation $y^2=x^2-x^3+x^4$ have solutions other than $x=1,y=1$? Interestingly, the Diophantine equation $y^2=x^2-x^3+x^5$ has such solutions: $x=3,y=15$, $x=5,y=55$, $x=56,y=23464$. However I don't know if there are any other solutions. The solution $x=3,y=15$ is related to the Mordell's equations $y^3=x^2+2$ and $y^3=x^2+18$ which have well known sole solutions $x=5,y=3$ and $x=3,y=3$. Indeed it follows from $3^3=5^2+2$ and $3^3=3^2+18$ that $3^5-(15)^2-(3^3-3^2)=0$.
The only integer solutions $(x,y)$ are $(0,0)$ and $(1,\pm 1)$.
Your equation is $y^2=x^2(1-x+x^2)$. This can only have integer solutions if $p(x)=1-x+x^2$ is a square or $y=0$. For $x>1$ we have $(x-1)^2<p(x)<x^2$, so $p(x)$ is not a square. For $x<0$ we have $x^2<p(x)<(x-1)^2$, so $p(x)$ is not a square. The only possibilities are $x=0$ and $x=1$, and they indeed give solutions.