It's the year 2017

Note that $2017$ is a prime number and $a$ and $b$ are positive integers satisfying the equation above

$\large \begin{cases} 28a^bb^a+1=2017 \\ a^b-b^a=1 \end{cases}$ or $\large \begin{cases} 28a^bb^a+1=1 \\ a^b-b^a=2017 \end{cases}$ (rej.)

$\large \Rightarrow \begin{cases} 28(1+b^a)b^a+1=2017 \\ a^b=1+b^a \end{cases}$

$\large \Rightarrow \begin{cases} (1+b^a)b^a=72 \\ a^b=1+b^a \end{cases}$

$\large \Rightarrow \begin{cases} b^a=8 \\ a^b=9 \end{cases}$

$\large \Rightarrow \begin{cases} a=3\\ b=2 \end{cases}$

$\large \Rightarrow a-b=3-2=1$