Power Diophantine equation involving primes: $(p+q)^q-p^q-q^q+1=n^{p-q}$ Suppose $p$ and $q$ are prime numbers, and $n>1$ is a positive integer. Find all solutions to the following Diophantine equation:$$(p+q)^q-p^q-q^q+1=n^{p-q}$$
What I have tried:
Obviously $p>q$. If $q=2$, we get one solution: $(p, q, n)=(3, 2, 13)$. From now on $p>q>2$. Letting $p-q=2^k r$, where $r$ is an odd number and working on the evaluation of $2$ in the equation gives $k=1$. 
I think that $r=1$ is the only possibility. But I don't know how to prove it! I tried to show that $r$ cannot have a prime divisor, but I failed!
Looking mod $p$, $q$ and $p+q$ gives $n^{p-q} \equiv 1 \pmod {pq^2(p+q)}$.
Edit: Let's put a new restriction on the equation: Order of $n$ modulo $p$ is $p-q$. If you work on equation with the new constraint, please inform me about your results!  
Notice that $n^{p-q} \equiv 1 \pmod p$, so $\gcd(n, p)=1$ and it follows from the Fermat's little theorem that $n^{p-1} \equiv 1 \pmod p$. Order of $n$ modulo $p$ is $p-q$, hence $p-q|p-1$ and consequently $p-q|q-1$.   
 A: $$(p+q)^q-p^q-q^q+1=n^{(p-q)}$$
Let $p-q=\alpha$
$$(p+q)^q-p^q-q^q=n^{\alpha}-1$$
Since q is odd, number of terms of expansion $(p+q)^q$ is even and terms $p^q$ and $q^q$ will be eliminated on LHS and the remained polynomial has a factor like $[p.q(p+q)]$ which is even.On RHS we have a factor like $n-1$ which indicates n must be odd, in fact we have:
$$p.q (p+q).P(p, q)= (n-1)(n^{\alpha -1}+n^{\alpha -2}+n^{\alpha -3}+ . . .)$$
Where $P(p, q)$ denotes a polynomial having parameters p and q. This can help us to choose the number for n to be checked. For example with $p=5 $ and $q=3$ we may write:
$$(5+3)^3-5^3-3^3=3(3\times 5)(3+5)=18\times 20=n^2-1=(n-1)(n+1)$$
So $q=19$ works.If $p=2k_1-1$ and $q=2k_2+1 $, we have:
$$C^q_r (2k_1-1)(2k_2+1)(2k_1+2k_2)=(n-1)\Sigma^{\alpha-1}_{r=1}n^{\alpha - r}$$
Now for selecting p and q we must consider that the difference between p and q must be such that we can construct a relation it's sides have close values, for example consider primes $p=47$ and $q=7$, we have:
$$(47+7)^7-47^7-7^7=n^7-1$$
Number of digits of $54^7$ is :
$N_d=7 \log 54  = 13$ so n must be such that it's 7th power has approximately 13 digits or we must have:
$\log n ≈ \frac{13}{7} $  ⇒ $n  ≈ 100$
Now on LHS we have:
$(2\times 47= 94)(26)(7)=4\times47 (7\times13=91) $
Numbers 91 and 94 are closed to 100, but n must be odd so number 95 can be selected for test. 
