9. The number of prime pairs $(a, b)$ that satisfy the equation
$$
a^{b} b^{a}=(2 a+b+1)(2 b+a+1)
$$

is $\qquad$.
9.2.

If $a=b$, then the equation is equivalent to $a^{2a}=(3a+1)^2$. But $a \mid a^{2a}, a \mid (3a+1)^2$, which is a contradiction, so $a \neq b$.
Assume without loss of generality that $a > b$. Since the difference between the two factors on the right side of the original equation is $a-b$, $a$ can only divide one of these factors, and $a^b$ also divides this factor.
Thus, $a^2 \leq a^b \leq 2a + b + 1 \leq 3a \Rightarrow a \leq 3$.
Therefore, $a=3, b=2$.
Verification shows that $(a, b)=(3,2)$ is a solution to the original equation.
By symmetry, $(a, b)=(2,3)$ is also a solution to the original equation.
In conclusion, the only solutions are $(3,2)$ and $(2,3)$.