In an article of Robert Friedman, I came up with a comment:
There are finitely many deformation types of Calabi-Yau threefolds for a given diffeomorhpic type if $b_2 =1$. And it is said that this is a special case of a result due to Kollar.
(p. 113, Friedman, Robert {On threefolds with trivial canonical bundle}. Complex geometry and Lie theory)
Can anyone explain why or give some references for it?
I am just posting my comment as an answer. The original form of Matsusaka's Big Theorem is that for a pair $(X,H)$ of a smooth, projective $r$-fold $X$ and an ample divisor $H$, there exists an integer $n_0$ that depends only on $r$ and the Hilbert polynomial $P(n) = \chi(X,\mathcal{O}_X(nH))$ such that for all $n\geq n_0$, the invertible sheaf $\mathcal{O}_X(nH)$ is very ample. The Kollár-Matsusaka Theorem proves that to find a uniform $n_0$, you do not need the entire Hilbert polynomial $P(n)$. You only need the leading coefficient, i.e., the intersection number $(H^r)_X$, and the next coefficient, or equivalently, the intersection numbers $(H^r)_X$ and $(H^r.K_X)_X$.
For a Calabi-Yau, $(H^r.K_X)_X$ equals $0$, so you only need $(H^r)_X$. Once you know that $\mathcal{O}_X(nH)$ is very ample, there is an old inequality (probably due to Castlenuovo) proved by taking hyperplane sections that gives a bound on the integer $m=h^0(X,\mathcal{O}_X(nH))$, namely $m \leq r + d$, where $d=n^r(H^r)_X$ is the degree of the projective embedding of $X$ in $\mathbb{P}^{m-1}_\mathbb{C}$ via the very ample invertible sheaf $\mathcal{O}_X(nH)$. Via the construction of the Hilbert scheme in Grothendieck's original Seminaire Bourbaki articles, or even just via the Chow variety, there is a quasi-compact, locally finite type scheme over $\mathbb{C}$ that parameterizes all smooth, degree $d$ closed subschemes $X\subset \mathbb{P}^{m-1}_{\mathbb{C}}$ that are purely $r$-dimensional. In summary, given an integer $v$, the family of smooth, polarized Calabi-Yau $r$-folds $(X,H)$ with $(H^r)_X\leq v$ is a bounded family.
For a smooth Kähler $r$-fold $X$ with $b_2(X)=1$, by the Hodge identities, $h^{2,0}(X)=h^{0,2}(X)=0$, or else $b_2(X) \geq h^{2,0}+h^{0,2} = 2h^{2,0}$ would be at least as large as $2$. Since $H^2(X,\mathcal{O}_X)$ equals $\{0\}$, the exponential exact sequence gives a surjection $$\text{Pic}(X)\twoheadrightarrow H^2(X,\mathbb{Z}(1)).$$ In particular, every integral cohomology class is the first Chern class of an invertible sheaf. Since $b_2(X)$ equals $1$, the torsion-free quotient of $H^2(X,\mathbb{Z})$ is isomorphic to $\mathbb{Z}$. Denote one generator by $g$.
By surjectivity of the Chern class map, there exists an ample divisor class $H$ on $X$ whose first Chern class maps to $\pm g$. Moreover, $H$ is unique up to adding divisor classes whose image in $H^2(X,\mathbb{Z})$ is a torsion class. The degree of the top self-cup product $\int_X g^r$ equals $\pm (H^r)_X$. This is insensitive to adding torsion-classes to $H$, since $H^{2r}(X,\mathbb{Z})$ is torsion-free. Thus, for one and every such ample divisor class $H$, the top self-intersection number $v=(H^r)_X$ is independent of the choice of ample divisor class, and equals the absolute value of the topological number $\int_X g^r$.
In this argument, it was not actually necessary that $X$ is simply connected. By the Universal Coefficients Theorem, the torsion in $H_1(X,\mathbb{Z})$ does give torsion in $H^2(X,\mathbb{Z})$, but the argument above uses only the torsion-free quotient of $H^2(X,\mathbb{Z})$.
