If $p^2+q^2+7pq = r^2$ ($r$ being any integer), then $(p+q)^2 + 5pq = r^2$. So $5pq = r^2-(p+q)^2 = (r+p+q)(r-p-q)$.
Since $p, q$ and $5$ are all prime, it follows that one of the factors on the right-hand side is equal to one of them, and the other factor is the product of the other two. As clearly visible, $r+p+q$ is greater than any of the numbers $p, q$ and $5$. So it must be the product of two of those numbers (maybe three too), and the other factor $r-p-q$ must be equal to $p, q$ or $5$ (or $1$).
Now, different cases arise:
CASE $1$:
If $r-p-q = p$ then $r=2p+q$, and thus original equation becomes $p^2+q^2+7pq = (2p+q)^2$, which simplifies to $p=q$, Same if $r-p-q = q$.
CASE $2$:
If $r-p-q = 5$ then $r = p+q+5$, and the equation $5pq = (r+p+q)(r-p-q)$ becomes $pq = 2(p+q)+5$. You can write this as $(p-2)(q-2)=9$, and the only solutions to this are $p=q=5$ and $p=3;q=11$ (or the other way round).
CASE $3$: $r+p+q$ might be equal to the product of all three numbers $5pq$, with $r-p-q = 1$. But then the equation becomes $2p+2q+1 = 5pq$, which is clearly impossible because the right-hand side is visibly greater than the left-hand side (Though there are solutions like $(1,1)$, but $1$ is not prime ).
So, to summarise the whole answer, We can say that only solutions are:
$(p,q)=(p,p),(3,11),(11,3)$
SOURCE
By Fermat's little theorem,
$$0\equiv n^7-n=n(n^3-1)(n^3+1)\pmod 7.$$
Because $7$ is prime, either $n\equiv0$, $n^3\equiv1$ or $n^3\equiv-1$, so the only possible cubic residues modulo $7$ are $-1,0,1$. These are all possible, as they are the residues of $(-1)^3,0^3,1^3$.
In general, things aren't that easy and one has to compute a list of residues mod $p$. Number theory tells us the following:
If $p$ is prime, then there are exactly $\frac{p-1}{\gcd(n,p-1)}$ nonzero $n$th power residues modulo $p$ (which, together with $0$ makes $1+\frac{p-1}{\gcd(n,p-1)}$).
In the case of $(p,n)=(7,3)$ as above, this is $1+\frac63=3$.
Best Answer
This solution is based on a solution to a similar problem about $2^n + 1$ found here.
No.
Suppose that $7^n + 1 = m^3$ for some positive integers $n$ and $m$. Rearranging, we get $$7^n = m^3 - 1 = (m-1)(m^2 + m + 1).$$
Because $7$ is prime, we know that both $m-1$ and $m^2 + m + 1$ must both be powers of $7$ (including $7^0 = 1$).
So, excluding $n = 1$, there is no other positive integers $n$ such that $7^n + 1$ is a perfect cube.