For variety, here's a method that doesn't require guessing at which small modulus to try. (however, it does use some more sophisticated mathematics)
First, we simplify our quadratic by completing the square $ (p+2q)^2 - 3q^2 = 11$
Next, we homogenize: $ (pr+2qr)^2 = 11 r^2 + 3 (qr)^2$
and for clarity, change variable $ z^2 = 11 x^2 + 3 y^2$ (rational solutions to the original equation would be $q = y/x$ and $p = z/x - 2q$)
Now, we invoke a two relevant sledgehammers.
This equation obviously a real solution, and the only relevant primes to check are $3$ and $11$.
$ (3, 11)_{3} = \left(\frac{11}{3}\right) = -1$ $ (3, 11)_{11} = \left(\frac{3}{11}\right) = 1$
(and from this, we can also conclude $(3,11)_2 = -1$) where the left hand side is the Hilbert symbol and the right hand side is the Legendre symbol.
Since we got a $-1$, the equation has no rational solutions, and thus it has no integer solutions.
Incidentally, this result explains why the other posts were successful when checking modulo $3$ and checking modulo $8$.
A lowbrow summary of this method would be
Complete the square, find a real solution, then try working modulo 8, then try working modulo every prime dividing one of the coefficients. If you found solutions in all cases, then your quadratic has rational solutions. Otherwise, it doesn't.
The main thing the sophisticated math tells you is that those are the only primes useful to check, and that if the check passes every test, then there truly are rational solutions. (actually proving none of the rational solutions are integer solutions is generally much trickier)