Proof For $n = 2^k \cdot m$ ($k$ is a non-negative integer, $m$ is odd), let $v(n) = 2^k$.
We prove by contradiction: assume $(x, y)$ is one positive integer solution of the equation. Let
$$v(x + i) = \max_{1 \le j \le 2014} \{v(x + j)\},$$
then if $1 \le j \le 2014,\ j \ne i$,
$$v(x + j) = v(x + i + (j - i)) = v(j - i),$$
so
$$v\left(\prod_{1 \le j \le 2014,\ j \ne i} (x + j)\right) = v\left((2014 - j)! \cdot (j - 1)!\right) \le v(2013!)$$
since $\prod_{j=1}^{2014} (x + j) = \prod_{j=1}^{4028} (y + j)$ is a multiple of $4028!$, thus
$$x + i \geq v(x + i) \geq v\left(\frac{4028!}{2013!}\right) > 2^{1007},$$
therefore $x > 2^{1006}$. So $(y + 4028)^{4028} > \prod_{j=1}^{4028} (y + j) = \prod_{j=1}^{2014} (x + j) > 2^{1006 \cdot 2014}$, we have $y + 4028 > 2^{503}$, $y > 2^{502}$.

Lemma: Let $0 \le x_i < \frac{1}{2}\ (1 \le i \le n)$, if $x = \frac{1}{n} \sum_{i=1}^{n} x_i$, $y = 2 \max_{1 \le i \le n} \{x_i^2\}$, then
$$1 - x \ge \left( \prod_{i=1}^{n} (1 - x_i) \right)^{\frac{1}{n}} \ge 1 - x - y.$$
Proof of lemma: By AM-GM inequality, the inequality on the left is easy to prove. The inequality on the right holds, since
$$\begin{aligned} \left( \prod_{i=1}^{n} (1-x_i) \right)^{\frac{1}{n}} &\ge \frac{n}{\sum_{i=1}^{n} \frac{1}{1-x_i}} \\ &\ge \frac{n}{\sum_{i=1}^{n} (1+x_i + 2x_i^2)} \\ &= \frac{n}{n+nx+2\sum_{i=1}^{n} x_i^2} \\ &\ge \frac{1}{1+x+y} \\ &\ge 1-x-y, \end{aligned}$$
the lemma is proved.

Back to the problem, Let $w = x + 2015 > 2^{1007}$, $z = y + \frac{4029}{2} > 2^{502}$, then the equation is equivalent to
$$w \cdot \left( \left(1 - \frac{1}{w}\right) \left(1 - \frac{2}{w}\right) \cdots \left(1 - \frac{2014}{w}\right) \right)^{\frac{1}{2014}} = z^2 \cdot \left( \left(1 - \frac{1}{4z^2}\right) \left(1 - \frac{9}{4z^2}\right) \cdots \left(1 - \frac{4027^2}{4z^2}\right) \right)^{\frac{1}{2014}}.$$
By lemma,
$$w\left(1 - \frac{2015}{2w}\right) > w \cdot \left(\left(1 - \frac{1}{w}\right)\left(1 - \frac{2}{w}\right)\cdots\left(1 - \frac{2014}{w}\right)\right)^{\frac{1}{2014}}$$
$$\begin{aligned} &> w \left( 1 - \frac{2015}{2w} - \frac{2 \cdot 2014^2}{w^2} \right) \\ &> w \left( 1 - \frac{2015}{2w} \right) - \frac{1}{8}, \end{aligned}$$
$$therefore the decimal of \$w \cdot \left( \left( 1 - \frac{1}{w} \right) \left( 1 - \frac{2}{w} \right) \cdots \left( 1 - \frac{2014}{w} \right) \right)^{\frac{1}{2014}}\$ belongs to \$\left( \frac{3}{8}, \frac{1}{2} \right)\$. On the other hand, by lemma$$
$$\begin{aligned} z^2 \left( 1 - \frac{1^2 + 3^2 + \cdots + 4027^2}{4z^2 \cdot 2014} \right) &> z^2 \cdot \left( \left( 1 - \frac{1}{4z^2} \right) \left( 1 - \frac{9}{4z^2} \right) \cdots \left( 1 - \frac{4027^2}{4z^2} \right) \right)^{\frac{1}{2014}} \\ &> z^2 \left( 1 - \frac{1^2 + 3^2 + \cdots + 4027^2}{4z^2 \cdot 2014} - \frac{2 \cdot 4027^4}{(4z^2)^2} \right), \end{aligned}$$
$$thus$$
$$\begin{aligned} z^2 - \frac{4 \cdot 2014^2 - 1}{12} &> z^2 \cdot \left( \left( 1 - \frac{1}{4z^2} \right) \left( 1 - \frac{9}{4z^2} \right) \cdots \left( 1 - \frac{4027^2}{4z^2} \right) \right)^{\frac{1}{2014}} \\ &> z^2 - \frac{4 \cdot 2014^2 - 1}{12} - \frac{4027^4}{8z^2} \\ &> z^2 - \frac{4 \cdot 2014^2 - 1}{12} - \frac{1}{8}. \end{aligned}$$
$$$$
Since $z^2 - \frac{4 \cdot 2014^2 - 1}{12}$ is an integer, so the decimal of $z^2 \cdot \left( \left( 1 - \frac{1}{4z^2} \right) \left( 1 - \frac{9}{4z^2} \right) \cdots \left( 1 - \frac{4027^2}{4z^2} \right) \right)^{\frac{1}{2014}}$ belongs to $\left( \frac{7}{8}, 1 \right)$, this is a contradiction.

Therefore, there are no 2-tuples $(x, y)$ of positive integers satisfying
$$\prod_{j=1}^{2014} (x + j) = \prod_{j=1}^{4028} (y + j).$$