Let $V_1$ be the defining 3-dimensional representation of SU(3) with character $\chi_1$. Likewise, let $V_2$ be the conjugate representation with character $\chi_2 = \overline{\chi_1}$. Then every irreducible representation of SU(3) has a character which is a $\mathbb{Z}-$polynomial in $\chi_1$ and $\chi_2$.

What does this have to do with your problem? Well, letting $u=\chi_1$ and $v=\chi_2$, your polynomials are simple linear combinations of the polynomials of the irreducible characters of SU(3). For example the polynomial $v^2-u$ corresponds to the character of the irreducible representation which is the symmetric square of $V_2$, which is the (0,2)-representation. Similarly $v^3-2uv+1$ corresponds to the character of the (0,3)-representation which is the third symmetric power of $V_2$. (Here the $(a,b)$-representation is the representation with highest weight $a\omega_1 + b\omega_2$ with $\omega_1$ and $\omega_2$ the highest weights of $V_1$ and $V_2$ respectively).

Now since the character $\chi_{[a,b]}$ of the $(a,b)$-representation is the complex conjugate of the character of the (b,a)-representation, the polynomial $P_{a,b}(\chi_1,\chi_2)$ expressing the character of the $(a,b)$-representation satisfies

$P_{a,b}(\chi_1,\chi_2) = \overline{P_{a,b}(\chi_2,\chi_1)} = P_{b,a}(\chi_2,\chi_1)$

This implies the polynomials $P_{a,b}$ satisfy your condition (if you want a short argument of this fact, give me a little while to think of something coherent).

But the short of it is, your polynomials correspond to characters of SU(3).

**Edit:** Inspired by Johann's and comment about the recurrence, I looked more closely at the polynomials you listed and realized all of them $are$ in fact polynomials corresponding to the (0,k)-representations of SU(3) (for k=2..8) which is also the irreducible representation $Sym^k(V_2)$. In particular I am pretty sure one can extend this to say that $any$ polynomial corresponding to an irreducible representation of SU(3) will satisfy the property of being 'remarkability'. I think therefore that the next logical step is to look for polynomials which do not correspond to irreducible representations (for example, take simple linear combinations of the above polynomials) and see if they too have the 'remarkable' property.

**Edit 2:** Just realized that the above statement cannot be entirely true as I have already pointed out in a comment to Johann's answer that $q = uv-1$ is the polynomial associated to the 8-dimensional irreducible adjoint representation; clearly $q(u,v) = q(v,u) = 0$ has infinitely many solutions $(u,v)$, all of the form $(x,\overline{x})$. This is the (1,1)-representation of SU(3), and similarly for every $k$ one has $P_{k,k}(u,v)$ is symmetric with respect to $u$ and $v$; aside from these polynomials I don't see any obvious obstruction to a general $P_{a,b}(u,v)$ satisfying your conditions as long as $a\neq b$.

**Edit 3:** Last edit I promise. One thing I overlooked in the previous edit; namely that $P_{k,k}$ divides $P_{a,b}$ whenever $a\equiv b\equiv -1\mod k+1$; in particular, aside from the $P_{k,k}$, there are a lot of other $P_{a,b}$ which do not have the property of remarkability. In such cases the structure of the cofactor $P_{a,b}/P_{k,k}$ is actually pretty well-behaved, and based on what I know about this structure it ought to be true that if $k+1 = gcd(a+1,b+1)$ then this cofactor has the property of remarkability. Some simple examples of such cofactors are $u^2-2v$ and $u^3-3uv+3$, so you might test these examples too (unfortunately I don't currently have access to Maple or Mathematica to do it myself).