Can there be system of irreducible polynomial equations over $\mathbb{Q}$ which has more than 5 but less than 50 $\mathbb{Q}$-solutions?
I somehow feel that I have to use the fact that polynomials in more than one variables over an infinite field have infinity many solutions. I'm thinking in terms of algebraic sets but I really can't come up with examples (or counterexamples).
For any $a \in \mathbb{Q}^n$, define $P_a(x_1,\ldots,x_n)=(x_1-a_1)^2+\ldots+(x_n-a_n)^2$. Given a finite subset $S \subset \mathbb{Q}^n$, let $P_S=\prod_{a \in S}{P_a}$. Then the zeroes of $P_S$ are exactly the roots of $S$.
Edit: I was recalled that the question was about irreducible polynomials. Then, natural objects to consider are the equations of “elliptic curves” ($y^2=x^3+Ax+B$ with $x^3+Ax+B$ separable as well) and “hyperelliptic curves” ($y^2=P(x)$, $P$ separable of degree at most five).
For elliptic curves, it is a well-known result of Mazur that, when this equation has finitely many solutions, its number of solutions is $15$, or $11$, or at most $9$ (and all cases occur). See for instance https://en.m.wikipedia.org/wiki/Torsion_conjecture .
For hyperelliptic curves (or any curve really but the formulation is a bit trickier), it is a theorem of Faltings (again, an important and difficult one) that the number of rational solutions is finite – and a difficult problem to provide in general a somewhat tight upper bound of the number of solutions (see however the very striking Theorem 1.1 of https://arxiv.org/pdf/2001.10276.pdf ).
Here is an explicit equation (a hyperelliptic curve): $y^2=x(x-1)(x-2)(x-5)(x-6)$ (from http://www-math.mit.edu/~poonen/papers/chabauty.pdf ). It has exactly nine solutions, that are: the five where $y=0$, $(3,\pm 6)$ and $(10,\pm 120)$. The proof in the reference that these are the only solutions is somewhat involved though.