14. Find all positive integer pairs $(x, y)$ that satisfy the equation $y^{x^{\prime}}=x^{y+2}$.

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14. Solution: Let the given equation be (1). Clearly, if one of $x, y$ is 1, then $(x, y)=(1,1)$.
Now assume $x>1, y>1$.
From the given condition, $x$ and $y$ have the same prime factors. Let $x=\prod_{i=1}^{n} p_{i}^{a_{i}}, y=\prod_{i=1}^{n} p_{i}^{2}$, then from (1) we have,
$$
x^{2} \beta_{i}=(y+2) \alpha_{i}, i=1,2, \cdots, n \text {. }
$$

If one of $p_{i}(i=1,2, \cdots, n)$ is odd, let it be $p_{j}$, then $\left(p_{j}, y+2\right)=1$.
$\because p_{j}^{a}\left|x, p_{j}^{2 a_{j}}\right| x^{2}$, and from (2) we get, $p_{j}^{2 a_{j}} \mid(y+2) \alpha_{i}$, $1 \leqslant i \leqslant n, \therefore p_{j}^{2 a_{j}} \mid \alpha_{i}, 1 \leqslant i \leqslant n$, in particular, $9^{a_{j}} \leqslant p_{j}^{2 a_{j}} \mid \alpha_{j}$. However, it is easy to prove that the inequality $9^{n}>n$ holds for any positive integer, so, $9^{a_{j}} \leqslant p_{j}^{2 a_{j}} \mid \alpha_{j}$ does not hold. Therefore, the prime factors of $x$ and $y$ can only be 2.

Let $x=2^{\alpha}, y=2^{\beta}(\alpha \geqslant 1, \beta \geqslant 1)$. From (1) we know, $2^{2 \alpha-1} \beta$ $=\alpha\left(2^{\beta-1}+1\right)$, then $2^{2 \alpha-1} \mid \alpha\left(2^{\beta-1}+1\right)$, since $2^{2 \alpha-1}>\alpha$ $\left(\forall \alpha \in \mathbf{N}^{*}\right)$, so $\beta=1$ (otherwise $2^{\beta-1}+1$ is odd). Thus, $2^{2 \alpha-2} \mid \alpha$. Using induction, it can be proven that when $n \geqslant 2$, $2^{2 n-2}>$ $n$, therefore, $\alpha=1, x=y=2$. The pairs of positive integers that satisfy the condition are (1, 1), $(2,2)$.