Cusofay wrote:

We can check that $P(x)=mx^4+nx^2$ for real numbers $m,n$ works and we want to prove its uniqueness.
Note that if $P$ and $Q$ are solutions, $aP+bQ$ is a solution for all real numbers $a,b$ so it's enough to find all solutions of the form $x^n$ . $P(0,0,0)$ and $P(0,0,x)$ gives us that $n$ is even

$P(6,3,-2) \Rightarrow 3^n+5^n+(-8)^n=2 \cdot 7^n$
If $n=0$ , then this obviously doesn't hold.

If $n=2$ , then $9+25+64= 98 =2 \cdot 49$ , which works.

If $n=4$ , then $81+625+4096 = 4802 = 2 \cdot 2401$ , which also works.

If $n=6$ , then $3^6+5^6+8^6>8^6=(7+1)^6=7^6+\frac{6}{7} \cdot 7^6 + \frac{15}{49} \cdot 7^6 = (2+\frac{8}{49}) \cdot 7^6>2 \cdot 7^6,$ which is a contradiction.

If $n \ge 8$ , then $3^n+5^n+8^n>8^n=(7+1)^n>7^n+n \cdot 7^{n-1} > 2 \cdot 7^n,$ contradiction.

$\mathbb{Q.E.D.}$

unfortunately, the linear combination of any two solutions also being a solution does not necessarily imply that checking $x^n$ suffices

for example, if hypothetically the solution set is $r(x+1)$ for real numbers $r$ , then it is true that the linear combination of any two solutions is still a solution, but in this case there are no solutions of the form $x^n$ even though there does exist solutions.

the issue here is that $1,x,x^2,\dots$ is not the only basis for real polynomials, and there are plenty of other choices, so closure under addition like this does not imply that only checking these suffices

it's ok though

I wish you good luck in the future!