$\underline{\textbf{Solution 1}}$

Let $A = \alpha$ and $B=2A=2\alpha$. Then by the Law of Sines: $$\frac{a}{\sin\alpha} = \frac{b}{\sin 2\alpha}\implies\frac{a}{\sin\alpha}=\frac{b}{2\sin\alpha\cos\alpha}\implies \cos\alpha =\frac{b}{2a}$$ Meanwhile, by Law of Cosines, we have $$\cos\alpha=\cos A =\frac{b^2 + c^2 - a^2}{2bc}\implies \frac{b}{2a}=\frac{b^2 + c^2 -a^2}{2bc}$$ Rearranging and factorizing this relation give \begin{align} b^2c &= a(b^2+c^2-a^2)\\ b^2c &=a(b^2-a^2 -ac) +ac^2 +a^2c\\ b^2c-ac^2-a^2c&=a(b^2-a^2-ac)\\ c(b^2 -ac - a^2) &=a(b^2-a^2-ac)\\ (c-a)(b^2-ac-a^2 )&=0 \end{align} Because $a$, $b$, and $c$ are consecutive integers, $c\ne a$. This means that \begin{equation} b^2 - ac - a^2 =0 \end{equation} Meanwhile, because $B > A$, it must have $b > a$. It follows that there are three possible scenarios: i) $c > b > a \implies c = a + 2, b = a + 1$ ii) $b > c > a \implies b=a+2, c= a+1$ iii) $b > a > c \implies b= a+1, c=a -1$ When $c = a + 2$ and $b = a + 1$, then $$ (a+1)^2 -a(a+2)-a^2 = 0 \implies a=\pm 1$$ Discarding negative value gives $a=1$ which implies $b=2$ and $c=3$. However, in this case, $\triangle{ABC}$ will become degenerated. When $b=a+2$ and $c= a+1$, then $$(a+2)^2 - a(a+1)-a^2=0\implies a = -1, 4$$ Discarding negative value gives $a=4$ which gives $b=6$ and $c=5$. When $b= a+1$ and $c=a -1$, then $$(a+1)^2 - a (a-1) -a^2=0\implies a=\frac{1}{2}\cdot(3\pm\sqrt{13})$$ Neither of them is integer. Therefore, there is only one triangle satisfying the requirement. The three sides of this triangle is $4$, $5$ and $6$.

$\underline{\textbf{Solution 2}}$

This problem can also be solved without using trigonometry.

As shown above, in $\triangle{ABC}$, let $\angle{B}=2\angle{A}$ and $BD$ be the angle bisector of $\angle{B}$. Note that $x+y=b$. Then $$\triangle{ABC}\sim\triangle{BDC}\implies\frac{AB}{BD}=\frac{BC}{DC}=\frac{AC}{BC}$$ $$\therefore\quad \frac{c}{y}=\frac{a}{x}=\frac{x+y}{a}$$ $$\implies\quad ac=y(x+y)\quad\text{and}\quad a^2=x(x+y)$$ Adding these two relations and replace $x+y$ with $b$ leads to $$ac + a^2 = b^2$$ Then, this equation can be analyzed using the same casework technique as the one employed in the original solution to find the final result.