Since the equation $p^2+7pq+q^2=n^2$ is symmetric in $p$ and $q$, we may assume without loss of generality that $p\lt q$ (setting aside the obvious solutions with $p=q$). For reasons that will become clear in a moment, it's easy enough to check for solutions with $p\lt q\le7$ (there are none).
Multiplying both sides of the equation by $4$, it's possible to rewrite it as
Now the prime $q$ cannot divide both factors on the left hand side, since the sum of those two factors is congruent to $4p$ mod $q$. But $2p+7q-2n\lt9q\lt q^2$, so we must have $2p+7q+2n=aq^2$ and $2p+7q-2n=b$ with $ab=45$. Multiplying each of these by $a$ and summing (to eliminate $n$), we have
which solves to
(We can disregard the negative square root, since $q\gt7$.) This implies $ap+1$ is a square, which clearly implies $ap+1=k^2$ for some integer $k\ge2$ (since $ap\ge2$). But now the inequality $q\gt p$ becomes $7+2k\gt k^2-1$, or $(k-1)^2\lt9$. Thus $k\lt4$, so there are just two possibilities: $k=2$ or $k=3$, correpsponding to $ap+1=4$ and $ap+1=9$. But since $p$ is primes and $a$ is a divisor of $45$ (hence odd), the only possibility is $p=3$ and $a=1$, which gives $q=7+2\sqrt4=11$. And indeed
And that's it: The only prime pairs $(p,q)$ for which $p^2+7pq+q^2$ is a square are $(3,11)$, $(11,3)$, and pairs of the form $(p,p)$.
Remark (added later, after a restless night's sleep): When I first wrote up this answer, I did the scratchwork to check that there are no solutions with $p\lt q\le7$. This did the job, to be sure, but it felt a little unsatisfying. It only later (in the middle of the night) occurred to me to think mod $8$: If $p\lt q$, then $q$, and hence $n$, are both odd, so that $q^2\equiv n^2\equiv1$ mod $8$, which reduces the equation to $p(p-q)\equiv0$ mod $8$. From this it's easy to see that $p\equiv q$ mod $8$, which implies $q\ge2+8=10$.