Set $y=x^2+6x+7,$ and note that the resulting system $$y^2+6y+7-x=0,\,\,\,x^2+6x+7-y=0$$ is symmetric. That is, performing the transformation $(x,y)\mapsto(y,x)$ leaves the system unchanged except for a permutation. Thus, for every solution $(x,y),$ it is always the case that $(y,x)$ is also a solution, where $x,y\in\mathrm R.$
It follows that this system can only have an even number of real solutions, which must be in pairs that are reflections of each other in the line $y=x,$ and they cannot both fall on this line because the given equation is a quartic with real coefficients, and as such cannot have an odd number of real roots.
But the parabolas defined by the system above are orthogonal, congruent parabolas. Thus, they can only intersect along the line $x=y.$ This contradicts the observation above that the points of intersection cannot fall on this line. Thus, there are no real solutions for the system, and therefore to the original quartic.