A conjecture about polynomials $P(x)$ and $Q(x)$ defining the same field is believed to be true, but unproven:
Let $P(x)$ be an $n$ degree polynomial.
There exists a polynomial $Q(x)$ of degree $n$, with fixed of terms from degrees $n$ to $(n/2)$ if $n$ is even, and $(n+1)/2$ if $n$ is odd, and varied terms degrees from $(n/2)-1$ to $0$ if $n$ is even, and degrees from $((n+1)/2)-1$ to $0$ if $n$ is odd such that $Q(x)$ defines the same field as $P(x)$ and also infinitely many polynomials of this form, but does not have the same discriminant as $P(x)$. Note that the fixed coefficients in $Q(x)$ do not have to necessarily match those in $P(x)$.
Example of conjecture (assuming true):
Let $P(x)$ $=$ $x^3-16x^2+8x+1$, since $P(x)$ is a $3$ degree, or cubic polynomial, the first $2$ coefficients can be fixed in $Q(x)$, and there exists infinitely solutions $(a, b)$ such that $P(x)$ and $Q(x)$ define the same field where:
$P(x)$ $=$ $x^3-16x^2+8x+1$
$Q(x)$ $=$ $4x^3+7x^2+ax+b$
Note the coefficients in any such $Q(x)$ may be chosen at random, and infinitely many polynomials of this form $Q(x)$ would define the same field $P(x)$.
Second, can someone please explain an algorithm for solving the example problem assuming conjecture is true?
Thanks for looking into this.
A counterexample is defining the field extension $\mathbf{F}_{16} / \mathbf{F}_2$. The only irreducible quartics over $\mathbf{F}_2$ are:
In particular, this field extension cannot be defined by a polynomial of the form
$$ x^4 + x^2 + \ldots $$
(I conjecture that this counterexample can be adapted to the $2$-adics, to provide characteristic zero counterexamples)
Furthermore, any finite field example would give counterexamples to the assertion that there are infinitely many polynomials of a given form.