**EDIT 2016.08.25** So I paid more attention to the conditions,
fixed some bugs, and found solutions more in line with the
problem. (However, I have not added code to guarantee coprimality
between any two of $x$, $y$, and $p$ yet.)
In particular, one must have $p = 3 \bmod 4$ and
$a+b$ is odd. Using small primes up to 400000 for trial
factorization and testing exponents up to 60, as well as ensuring that both
factors were even, I found factorizations for $p^{a+b} + 1$
where both factors were even and the easier one was $-1 \bmod p$,
for primes 3,11,19,43,59,67,83,107,131,139,163,179, and others.
Possibly there are exponents for bases 7,23,31,47,71,79, etc.,
but my search was rather limited. $a+b=9$ worked for many $p$,
including 11,43,59,67, and 179. I am still confident there are
many solutions for $a=1$.

For $a=2$ I found solutions for 3,11,19,59, 67, and other primes .
I will report on $a=3$ later in a comment. I apologize for the goof below. While I still believe there are infinitely many solutions for a fixed b, my evidence is weaker than before. I think $a+b=45$ will have solutions for many primes. **END EDIT 2016.08.25**

I decided to run a computer program to convince myself of the paucity of examples, without fixing $b$. So, for many small odd primes $p$ and small exponents $c$, partially factor $p^c+1$ and stop when you find a divisor which is $-1 \bmod p$ for some $c$. For simplicity I am doing trial division with small primes.

Well. Just trial dividing with primes below $60000$, and exponents below $100$, I am finding example exponents with such divisors for every odd prime base $\lt 1000$ except 503, 719, and 823. If I did trial factorization with more primes I might find exponents for those bases also. So for $a=1$ I believe there are not only finitely many examples. Since an example for $a \gt 1$ is also an example for $a=1$, I think you have something confused in the post at the time of this writing. Of course, I am not fixing $b$ as you are, but still.

Of course, if $p^e+1$ has such a factor congruent to $-1 \bmod p^a$, then so does $p^{ek} +1$ for any odd positive integer $k$, and the question now becomes how big do you want $a$ to be. Running the program for $a=2$ I get no small examples for 31,71,97 and more primes, but it looks like I could fix that by improving the trial divisor bound. For $a=3$ my limited search does not find examples with the base being 13 or 19 or lots of other primes; if you have more CPU cycles you can probably find examples for those bases.

If you have fixed $b$, note that algebraic factorization can play a role especially when $a+b$ is 'more composite than usual'. Unless you have further specifications on $b$, my guess is you can find plenty
of $p$ with the right choice of $a$.

Gerhard "Maybe Factoral Abundance Helps Here" Paseman, 2016.08.25.