One possible solution
$2112$
Reasoning
Without loss of generality, suppose $U < V$, then we can simplify the expression to $$2(PQ^U)(PQ^{V-U}+1)$$ Given that we are looking for a "year" and that this is also even, it is likely that the year begins with $2$.
Then, we can look at the simplest case where $U=V-U=1$ and the expression becomes $$2(PQ)(PQ-1)$$ If we start with $2002$, we see that this factorises as $2 \times 7 \times 11 \times 13$ and can determine relatively quickly that this cannot match the above expression.
Continuing on with $2112$, we see that this factorises as $2^6 \times 3 \times 11 = 2 \times 2^5 \times 3 \times 11 = 2 \times 32 \times 33$ which is conveniently in the form above.
Hence we have $$PQ = 32 \,\,\,\,,\,\,\,\, U=1\,\,\,\,,\,\,\,\, V=2$$
Two other palindromic properties:
1. If we write $2112$ on a digital display and rotate $180^o$, the result is again $2112$.
2. The number $2112$ can be expressed as a palindromic product, that is $$2112 = 2 \times 33 \times 32 = (2)(33)(32)$$