Suppose that the prime factors of $n^2+1$ are all bounded by $N$ for infinitely many $n$. Then infinitely many integers $n^2 + 1$ can be written in the form $D y^3$ for one of finitely many $D$. Explicitly, the set of $D$ can be taken to be the finitely many integers whose prime divisors are all less than $N$, and whose exponents are at most $2$. For example, if $N=3$, then $D \in \{1,2,4,3,6,12,9,18,36\}$. Letting $x = n$, it follows that for at least one of those $D$, there are infinitely many solutions to the equation $x^2 - D y^3 = -1.$ From your original post, I'm guessing you actually know this argument, except you converted $n^2 + 1$ to $D y^2$ rather than $D y^3$ (of course, one can also use $D y^k$ for any fixed $k$, at the cost of increasing the number of possible $D$).
It turns out, however, that the equation $x^2 - D y^3 = -1$ only has finitely many solutions, and that this is a well known consequence of Siegel's theorem (1929), which says that any curve of genus at least one has only finitely many integral points. Siegel's proof does indeed use the Thue-Siegel method, although the proof is quite complicated. It is quite possible that this is the argument that Chowla had in mind - is is certainly consistent, since Chowla's paper is from 1934 > 1929.
There are some more direct applications of Thue-Siegel to diophantine equations, in particular, to the so called Thue equations, which look like $F(x,y) = k$ for some irreducible homogeneous polynomial $F$ of degree at least three. A typical example would be: $x^n - D y^n = -1,$ for $n \ge 3$. Here the point is that the rational approximations $x/y$ to $\sqrt[n]{D}$ are of the order $1/y^n$, which contradict Thue's bounds as long as $n \ge 3$. Equations of this kind are what are being referred to in the wikipedia article.
Edit Glancing at that paper of Chowla in your comment, one can see the more elementary approach. Let $K$ denote the field $\mathbf{Q}(i)$, and let $\mathcal{O} = \mathbf{Z}[i]$ denote the ring of integers of $K$. Assume, as above, that there exists an infinite set $\Sigma$ of integers such that $n^2+1$ has prime factors less than $N$. For $n \in \Sigma$, write $A = n+i$ and $B = n-i$; they have small factors in $\mathcal{O}$ (which is a PID, although the argument can be made to work more generally using the class number). As above, one may write $A =(a + bi)(x + i y)^3$ where $a + bi$ comes from a finite list of elements of $\mathcal{O}$ (explicitly, the elements whose prime factorization in $\mathcal{O}$ only contains primes dividing $N$ with exponent at most $2$). Since $\Sigma$ is infinite, there are thus infinitely many solutions for some fixed $a + b i$, or equivalently, infinitely many solutions to the equations (taking the conjugate): $n + i = (a + b i)(x + i y)^3, \qquad n - i = (a - b i)(x - i y)^3,$ Take the difference of these equations and divide by $2i$. We end up with infinitely many solutions to the equation: $b x^3 + 3 a x^2 y - 3 b x y^2 - a y^3 = 1.$ This is now homogeneous, so one can apply Thue's theorem, rather than Siegel's Theorem. (Explicitly, the fractions $x/y$ are producing rational approximations to the root of $b t^3 + 3 a t^2 - 3 b t - a = 0$ which contradict the Thue bounds.)