$p^{p+1}+(p+1)^p-1$ a perfect square For which primes $p$ is $$p^{p+1}+(p+1)^p-1$$ a perfect square? 
Context:
Once again a modified problem, this time from $p^{p+1}+(p+1)^p$ a perfect square, for which the answer is no such $p$ exist.  For the modified form above, however, $p=2,3$ work, the main trouble I've had being that one can't easily find prime divisors of the above number that aren't already divisors of $p$ or $p+1$.
Edit:
I forgot to say that I have already tested this statement up to $p=19$ or as far as the calculators on a computer will go without finding any other examples apart from $2$ and $3$
 A: This is my experience :
$A=P^{p+1}+(p+1)^p -1$
$p^{p+1}=(p^2)^\frac{p+1}{2}=k_1 p^2$
$(p+1)^p=∑  C^p_r p^r +1= k_2 p^2 +1$
$⇒A=P^{p+1}+(p+1)^p -1=(k_1+k_2)p^2 $
Also:
$P^{p+1}=(p+1-1)^{p+1}=∑ C^{p+1}_r (p+1)^r +1=t_1 (p+1)^2+1$
$(p+1)^p = (p+1)^2 (p+1)^{p-2}= t_2(p+1)^2$
$⇒A=P^{p+1}+(p+1)^p -1=(t_1+t_2)(p+1)^2 $
$⇒A=P^{p+1}+(p+1)^p -1=m p^2(p+1)^2 $
Where m is a composite integer; for example:
$7^8+8^7-1=23 . 109 . 56^2$
It can be seen that for Fermat primes a power of $2$  is also among the factors of m:
$(2^{2^n}+1)^{2^{2^n}+2}+(2^{2^n}+2)^{2^{2^n}+1}-1= k . (2^{2^n}+1)^2. (2^{2^n}+2)^2.2^{2^n}$
For example:
$17^{18} +18^{17}-1= k .2^4 . 17^2 . 18^2 $
There may exist primes of form  $(x^{2^n}-1)x^{2^n}+1$ which result in A such that a composite factor like $m$, composed of even powers of some prime factors of $x$, arises; for example:
$p=15 . 16 + 1= 241$;  $ 241^{241}+242^{241}-1=k . 2^4 . 241^2 . 242^2$
Conclusion:
We may conclude that;
a : The solutions are only $p=2$ and $p=3$ because m has square free prime factors.
b: Conjecture: There are primes other than 2 and 3 so that $A=P^{p+1}+(p+1)^p -1$ is perfect square.
