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x=y=1

and x=0 is trivial
sir what is your method

I do not have one yet. I can show with Mathematica than no
1

np sir :) will work on this later

clearly \(x\ge0 \)
for x=0 we have 2 solution (x,y)=(0,0),(0,-2)
x>0\[3^{2x}-3^x=y^4+2y^3+y^2+2y\]\[4(3^{2x}-3^x)+1=(2\times3^x-1)^2=4(y^4+2y^3+y^2+2y)+1\]so\[z=4(y^4+2y^3+y^2+2y)+1=4y^4+8y^3+4y^2+8y+1\]must be a complete square but\[(2y^{2}+2y)^{2}