prove that the system
(x^6)+(x^3)+(x^3)y)+y= (147^157)
(x^3)+((x^3)y)+(y^2)+y+(z^9)=(157^147)
has no solutions in integers x, y and z.
$\displaystyle x^6+x^3+x^3y+y= 147^{157}$
$\displaystyle x^3+x^3y+y^2+y+z^9=157^{147}$
The first equation $\displaystyle (x^3)^2+(y+1)x^3+(y-147^{157})=0$ is quadratic, so $\displaystyle x^3=\frac12(-(y+1)\pm\sqrt{(y+1)^2+4y-4*147^{157}})$
So $\displaystyle x^3$ can be substituted in the second equation $\displaystyle x^3(y+1)+y^2+y+z^9=157^{147}$. After simplifying, we arrive at...
$\displaystyle 4\left(z^9+\frac12(y^2-1-2*157^{147})\right)^2=8y^3+(17-4*147^{157}+2*157^{147})y^2+$ $\displaystyle (6-8*147^{157})y-2(2*147^{157}+2*(157^{147})^2+3*157^{147}-1)$
(Someone may want to check my algebra for small errors.)
Since the LHS is a perfect square, it would be sufficient to show that no integer values of y allow the RHS to be a perfect square. One approach may be to attempt to factor the RHS as a perfect square minus a small constant, and then show that there are only a finite number of squares who differ by that constant -- the smaller the better. This may be implausible by hand.
Hope this brute force progress helps!