A quantitative example of applications of linear forms in logarithms is the following result from [S.D. Adhikari, N. Saradha, T.N. Shorey, and R.Tijdeman, Transcendental infinite sums, Indag. Math. (N.S.) 12:1 (2001) 1--14; Theorem 4 and Corollary 4.1] which is cited in many other articles.
Theorem. Let $P(x)$ and $Q(x)$ be two polynomials with algebraic coefficients such that $Q(x)$ has simple rational zeros and no others. Let $\alpha$ be an algebraic number. Then, assuming the convergence of the series
$$
S=\sum_{n=1}^\infty\frac{P(n)}{Q(n)}\alpha^n,
$$
the number $S$ defined by it is either rational or transcendental. Furthermore, if all zeros of $Q(x)$ lie in $-1\le x<0$, then either $S=0$ or $S$ is transcendental.
The theorem gives an elegant criterion for deciding whether a number of this particular form is transcendental or not (like continued fractions allow us to decide whether a given number is from a quadratic field or not). However it is not applicable for the sums like
$$
\zeta(3)=\sum_{n=1}\frac1{n^3},
$$
as factorisation of the denominator polynomial, $Q(x)=x^3$, involves multiple rational zeros.
$\newcommand\p{\mathfrak{p}}$
$\newcommand\OL{\mathcal{O}}$
$\newcommand\P{\mathfrak{P}}$
Here is a solution which is essentially an elaboration on Felipe's answer.
Instead of working with squares, consider working with $m$th powers instead.
Lemma: If $(1 - v u^m)$ is exactly divisible by a prime $\p$ of $\OL_K$, then, assuming $K(v^{1/m}) \ne K$, the prime $\p$ is not inert in $L = K(v^{1/m})$.
Proof: In $\OL_L$, the ideal $\P = (\p,1 - v^{1/m} u)$ has norm $\p$.
By Cebotarev, the density of primes $\p$ which remain inert in $\OL_L$ is non-zero. Hence by the analytic
arguments Felipe alluded to, the set of principal ideals of the form $(1 - v u^m)$ with $v$ ranging over a the (finite) set of non-zero representatives in $\OL^{\times}_K/\OL^{\times m}_K$ has density zero. So it remains to deal with ideals of the form $(1 - u^m)$, where we now have flexibility in choosing $m$.
Lemma: Let $\ell$ be any prime. Suppose that $m = |(\OL_K/\ell)^{\times}|$. Then the density of principal ideals of the form $(1 - u^m)$ is at most $1/\ell$.
Proof: Since $u^m \equiv 1 \mod \ell$, this is the same as saying that the density of principal ideals divisible by $\ell$ is at most $1/\ell$.
Taken together, it follows that the density of principal ideals of the form $(1 - u)$ has density at most $1/\ell$ for any prime $\ell$, and hence has density zero.
Best Answer
There is a big difference between linear forms in many logarithms and in two (or three) logarithms. The first case is covered in the archimedean case by the work of E. Matveev; Matveev's original works are hard even to specialists but there is a very nice survey [Yu. Nesterenko, Linear forms in logarithms of rational numbers, in Diophantine approximation (Cetraro, 2000), 53--106, Lecture Notes in Math., 1819, Springer, Berlin, 2003. MR2009829 (2004i:11082)]. The $p$-adic case was mostly done by Kunrui Yu. The best estimate for the case of two logarithms, which is of importance because of Tijdeman's application to Catalan's equation, is given in [M. Laurent, M. Mignotte et Y. Nesterenko, Formes linéaires en deux logarithmes et déterminants d'interpolation, J. Number Theory 55 (1995), no. 2, 285--321. MR1366574 (96h:11073)]. The latest news in the last direction (also in relation to Catalan's) are reviewed in [M. Mignotte, Linear forms in two and three logarithms and interpolation determinants. Diophantine equations, 151--166, Tata Inst. Fund. Res. Stud. Math., 20, Tata Inst. Fund. Res., Mumbai, 2008. MR1500224 (2010h:11119)]