Almost Taxicab, Almost FLT

Let $m$ be a non-negative integer. Set $x = m(2m^2 + 1)\\y = -m(2m^2 + 1)\\z = 2m^2 + 1$ Then $x^2 + y^2 + z^2 \\= m^2(2m^2 + 1)^2 + m^2(2m^2 + 1)^2 + (2m^2+1)^2 \\= 2m^2(2m^2 + 1)^2 + (2m^2 + 1)^2 \\= (2m^2 + 1)(2m^2 + 1)^2 \\= (2m^2 + 1)^3 \\= z^3 \\= x^3 + y^3 + z^3.$

Where the last equality follows from $x = - y$ implying $x^3 = -y^3$ . Since $m$ can take any non-negative integer value, we may construct infinitely many triples $(x, y, z)$

First, let z = -x so that we are left with only one variable on RHS.

Therefore, ${x}^{2} + {y}^{2} + {x}^{2} = {y}^{3}$

2 ${x}^{2}$ + ${y}^{2}$ = ${y}^{3}$

2 ${x}^{2}$ = ${y}^{2}(y -1)$

${x}^{2}$ = ${y}^{2}(\frac{y-1}{2})$

$\frac{y-1}{2}$ has to be ${t}^{2}$

Hence, y = 2 ${t}^{2}$ + 1 = y for all integers t.

So, there are infinitely many solutions for this equation.

You can also substitute x or y instead of z to prove it.

So this is another question in form of solution. Actually, my solution isn't a good one. Where did I go wrong?

By rearranging we get, $x^{2}(x-1)+y^{2}(y-1)+z^{2}(z-1)=0$ Let's consider $f(x)= x^{2}(x-1)=0$

Range of $f(x)$ is $(- \infty,+ \infty)$ So, for any $x$ we have $f(x)$ such that there is always a $y$ for which $f(y) = - f(x)$ or $f(x) + f(y)=0$ . (Please trace the graph of $f(x)$ to have a clearer understanding of what I'm trying to say.) Let's say, we have $f(y) + f(x)=t$ such that $x , y$ are integers. Then $t$ must also be an integer. Let's suppose $f(z) =-t$ such that $z= \frac{p}{q}$ such that $p$ , $q$ are co-primes. Then, $f(z)$ can never be an integer, but this is contrary to the fact that $-t$ is an integer. Hence, $z$ also must be an integer that satisfies $f(z) =-t$ . Then, there is always an integer $z$ such that $f(z) =-t$ . Then, adding the above two equations, we have $f(x) + f(y) + f(z)=0$ . Hence, there are infinite solutions!