= Solution
Use the following fixed-number form of a logarithmic estimate: for chosen <logarithms> of fixed nonzero <algebraic numbers> $a_1,\ldots,a_s$, a nonzero form $\Lambda=\sum b_j\log a_j$ with <integer> coefficients $|b_j|\leq B$ satisfies $|\Lambda|\geq c(2B)^{-C}$, for fixed positive constants $c,C$. This follows from the <Baker lower bound for a homogeneous linear form in logarithms>.
By the <Dirichlet unit theorem>, write units as $x=\zeta\prod\epsilon_j^{m_j}$ and $y=\xi\prod\epsilon_j^{n_j}$, with finitely many <roots of unity> and fixed fundamental units. If the unit rank is zero, there are already only finitely many choices. Otherwise put $B=\max(1,|m_j|,|n_j|)$. The real logarithmic embedding of the free <unit group> has full rank and lies in the product-formula hyperplane. Equivalence of norms in that finite-dimensional space shows that a unit with exponent size $B$ has some archimedean conjugate of modulus at most $e^{-c_0B}$: a large positive logarithmic coordinate forces a comparably large negative one, and a large negative coordinate itself suffices.
Interchanging the two terms if necessary, choose such a conjugate for $y$. At this embedding the equation makes $\sigma(\alpha x)=1-\sigma(\beta y)$ with $|\sigma(\beta y)|\leq C_0e^{-c_0B}$. For large $B$ define the <logarithm> near one. It is nonzero because $\beta y\ne0$, and
$$
0<|\Lambda|=|\log(\sigma(\alpha x))|\leq2C_0e^{-c_0B}.
$$
On fixed branches it is
$$
\Lambda=\log\sigma(\alpha)+\log\sigma(\zeta)+\sum_jm_j\log\sigma(\epsilon_j)+2k\log(-1)
$$
for an <integer> $k=O(B)$. The bound on $k$ follows from summing the fixed imaginary parts and then reducing to the near-one branch. There are only finitely many possible embeddings and <roots of unity>, so the stated lower bound applies uniformly with coefficients $O(B)$. It contradicts the exponential upper bound when $B$ is large. Hence the exponents are bounded and \b[$\boxed{\alpha x+\beta y=1\text{ has finitely many unit solutions}}$] in $\mathcal O_K^\times$. This is the <unit equation over a number field> proof; “unit” here means an algebraic-<integer> unit, not an arbitrary nonzero field element.
For the <Thue theorem>, let $F\in\mathbb Z[X,Y]$ be irreducible homogeneous of degree $d\geq3$ and let $m\ne0$ be fixed. In a <splitting field> $L$, write $F(X,Y)=a\prod_i(X-\theta_iY)$ with distinct roots. Choose a fixed <integer> $c$ so that each $c\theta_i$ is integral and $c^dm/a$ is a nonzero <integer>. For a solution $(u,v)$ the nonzero <algebraic integers> $t_i=c(u-\theta_iv)$ satisfy $\prod_it_i=c^dm/a$. Their <principal ideals> therefore divide a fixed <ideal> and have only finitely many possibilities. For each occurring <principal ideal> choose a generator $\gamma_i$, so $t_i=\gamma_i\varepsilon_i$ with $\varepsilon_i$ a unit.
Any three distinct roots give the identity
$$
(\theta_2-\theta_3)t_1+(\theta_3-\theta_1)t_2+(\theta_1-\theta_2)t_3=0.
$$
Divide by the third term and absorb the finitely many generator ratios into the coefficients. This is a fixed-coefficient unit equation in $\varepsilon_1/\varepsilon_3$ and $\varepsilon_2/\varepsilon_3$, so it has finitely many solutions for each generator choice. Thus $\rho=t_1/t_3$ has finitely many values. The relation $(1-\rho)u=(\theta_1-\rho\theta_3)v$ fixes a rational direction whenever an <integer> solution exists; $\rho=1$ simply gives $v=0$. On each such direction, $(u,v)=k(u_0,v_0)$ for a primitive <integer> pair, and $k^dF(u_0,v_0)=m$ allows finitely many $k$. This proves the desired finiteness. When $m=0$, irreducibility excludes a nonzero rational root direction, leaving only $(0,0)$.
The argument also proves finiteness for any <binary form> with at least three distinct projective roots: irreducibility was used only to guarantee distinct roots and to dispose of the zero-value case. A rational invertible change of variables can move every projective root away from infinity; after clearing its fixed denominators, the resulting pairs lie in an <integer> lattice and the same proof applies. This stronger nonzero-value version will apply to the binomial forms in question 6.
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