Given an $n$ degree equation in 2 variables ($n$ is a natural number) $$a_0x^n+a_1x^{n-1}+a_2x^{n-2}+\cdots+a_{n-1}x+a_n=y^n$$
If all values of $a$ are given rational numbers, are there any known minimum or sufficient conditions for $x$ and $y$ to have:
solutions and how many of them would exist. If it is not known/possible (or too hard) for an $n$ degree polynomial, do such conditions exist for quadratic ($n=2$) and cubic ($n=3$) polynomials.
On
(The OP suggested a connection to this post.)
Because this question is too broad, it spreads thin and may be vague. I suggest it be limited so coefficients $a_i$ are rational, and $x,y$ are also rational.
Having said that, two nice results are discussed in Kevin Brown's website.
I. Deg 2: The sum of $24$ consecutive squares.
$$F(x) = x^2+(x+1)^2+(x+2)^2+\dots+(x+23)^2=y^2\tag1$$
$$F(x) = 24x^2+552x+4324=y^2$$
which has solution,
$$x=p^2+70pq+144q^2,\quad\quad y =10(7p^2+30pq+42q^2)$$
where $p,q$ solve the Pell equation $p^2-6q^2=1$. This has an infinite number of integer solutions with the case $p,q = 1,0$ yielding the famous cannonball stacking problem,
$$1^2+2^2+3^2+\dots+24^2 =70^2$$
II. Deg 3: The sum of $n$ consecutive cubes.
$$G(x) = x^3+(x+1)^3+(x+2)^3+\dots+(x+n-1)^3=y^3\tag2$$
$$G(x) = n x^3 + \tfrac{1}{4}n(n - 1)\big(6x^2 + 4 n x - 2x + n(n - 1)\big) =y^3$$
a solution of which (by Dave Rusin) is,
$$x=\tfrac{1}{6}(v^4 - 3v^3 - 2v^2 + 4),\quad\quad n=v^3$$
hence for $2^3=8$ and $4^3=64$ cubes,
$$(-2)^3+(-1)^3+\dots+3^3+4^3+5^3 = 6^3$$
$$6^3+7^3+8^3+9^3+10^3+\dots+69^3 = 180^3$$
and so on.
III. Deg 4: The sum of 4th powers in arithmetic progression.
No analogous results known so far. See linked post in first line.
On
You can determine whether a real solution to that equation exists; obviously, there is always one if $n$ is odd, since the left side, $y^n$ can be chosen, independent of the right side, to equal whatever we want. For even $n$, we want to know whether the left hand polynomial is ever positive - this clearly holds if $a_0$ is positive, since the function would tend to $\infty$ as $|x|$ gets large. However, if $a_0$ is negative, then the left hand function will be non-negative somewhere if and only if it has at least one real root. Thus, the problem would then collapse to determining if a univariate polynomial has a root and, in general, this is possible to accomplish computationally via Sturm sequences.
This addresses user45195's question and is too long for a comment.
When I said too broad, it was because the question originally didn't limit the field of $x$. A familiar field is the complex numbers $\mathbb{C}=a+bi$, of which a special case are the reals $\mathbb{R}$, and even more limited, the rationals $\mathbb{Q}$.
If $x$ was in $\mathbb{C}$, then it's just an old result (Fundamental Theorem of Algebra), that for any $y$, one can always find $n$ roots $x$ that solve the equation in your post, and there's nothing new to be said.
However, if $x,y$ are limited to the rationals $\mathbb{Q}$, that's where it gets interesting. The equation,
$$f(x) = y^2\tag1$$
where the degree $d$ of $f(x)$ is either $d = 2,3,4,5$ has been extensively studied. See algebraic curve, including Pell equations ($d=2$), elliptic curve ($d=3,4$), and hyperelliptic curve ($d=5$). For,
$$f(x) = y^3\tag2$$
then $d=3$ still has special cases as elliptic curves. For $d=4$, see trigonal curve. However, for the higher,
$$f(x) = y^m\tag3$$
where $d,m>3$, is more complicated. See superelliptic curve.