System of equations

I saw that

$x^2y+y^2z+z^2x=2186=2187-1=3^7-1$

$xy^2+yz^2+zx^2=2188=2187+1=3^7+1$

so I first to solve

$x^2y+y^2z+z^2x=xy^2+yz^2+zx^2=3^7$

I get $(9,9,9)$

Besides that, I found that , for the original question, it is obviously can't have $x=y=z$ because $2186\neq2188$

if I let $x=y$ , then $x^3+x^2z+xz^2=2186$ and $x^3+xz^2+x^2z=2188$ also leads to contradiction.

Same as when $y=z$ and $z=x$ .

So for the original question, $x\neq y \neq z$ .

I test $8,9,10$ try to get $3^7-1$ and $3^7+1$ by both equation, then it holds true!

$x^2+y^2+z^2=64+81+100=\boxed{245}$