How to disprove $8q^2+15=7m^2$ for any integral value of $q$ and $m$ 
Question Statement:-
Show that the quadratic equation $x^2+7x-14(q^2+1)=0$, where $q$ is an integer, has no integral roots.


My attempt at a solution:-
For the given quadratic equation having rational coefficients  to have integral roots the discriminant $D$ of the equation needs to be a perfect square.
$$\therefore D=(7)^2+56(q^2+1)=56q^2+105=7(8q^2+15)$$
Now, from the expression for the discriminant we see that $(8q^2+15)$ is always odd so for the discriminant to be a perfect square, the following condition needs to be satisfied
$$8q^2+15=7(2n+1)^2,\qquad\qquad\text{where n $\in Z$}$$
So $$2(q^2+1)=7(n^2+n)$$
After this I was having trouble coming up with a way to disprove the above equality that I came up with, if you could help me out as to how to proceed from here on or maybe suggest a way that makes it easier for me to prove what the question asks.

If $x^2+7x-14(q^2+1)=0$, where $x,q\in\mathbb Z$, then $$7\mid 7\left(2\left(q^2+1\right)-x\right)=x^2$$ so by Euclid's Lemma $7\mid x$, so $7^2\mid x^2$ and $7^2\mid 7x$, so $$7^2\mid x^2+7x=7\left(2\left(q^2+1\right)\right),$$
so $7\mid 2\left(q^2+1\right)$. Since $\gcd(7,2)=1$, by Euclid's Lemma $7\mid q^2+1$, i.e. $q^2\equiv -1\pmod{7}$, which is impossible, because $0^2\equiv 0$, $(\pm 1)^2\equiv 1$, $(\pm 2)^2\equiv 4$, $(\pm 3)\equiv 2$ mod $7$, so $-1$ isn't a quadratic residue mod $7$.
Another way here is noticing that $7$ is a prime of the form $4k+3$, $k\in\mathbb Z$, so by Quadratic Reciprocity (or a separate lemma, which is a lot easier to prove than quadratic reciprocity) $-1$ isn't a quadratic residue mod $7$.
