I am looking for the best known upperbound and lowerbound for $m$ in the equation $m^2 - v(z^2) = 1$ such that $(m,z)$ is the fundamental solution of the pell's equation $x^2 -v(y^2) = 1$.
I.e is it true that the fundamental solution $(m,z)$ satisfies $m < v!$
The fundamental solution of Pell's equation $x^2-Dy^2=\pm1$ can be computed from the continued fraction of $\sqrt{D}$ (see Yang, Theorem 2.6, and Mathworld), that by Lagrange's theorem has the following form: $$ \sqrt{D}=[a,\overline{c_1,c_2,\ldots,c_m,\ldots,c_2,c_1,2a}] \tag{1}$$ so that $$ \frac{x}{y}=[a,c_1,c_2,\ldots,c_m,\ldots,c_2,c_1] \tag{2} $$ It follows that the size of the fundamental solution depends on the length $\ell$ of the period of the continued fraction of $\sqrt{D}$. Since $c_i\leq a\leq\sqrt{D}$, a bound that is not far from being sharp in many cases is $x\leq D^{\ell/2}$. $\ell$ depends on the class number of $\mathbb{Q}(\sqrt{D})$, hence $$ x\leq D^{c\sqrt{D}} \tag{3}$$ for some explicit constant $c$ follows from Minkowski's bound. So we have that $x\leq D!$ holds for sure for any $D$ large enough.