Let us do a proof-by contradiction: Suppose $q = 1 +x, p = 1+x+y$ for non-negative integers $x,y$. Taking: $a = (p+\sqrt{p^2+q})^2 = 2p^2 + q + 2p\sqrt{p^2+q}$ (i); we require the radicand to be a perfect square so that $a \in \mathbb{N} \subset \mathbb{Q}$. Next, we take: $p^2+q = (1+x+y)^2 + (1+x) = x^2 + 2xy + y^2 + 2(x+y) + 1 + 1 + x = x^2 + 2xy + y^2 + 3x+2y + 2 = K^2$ (ii); for $K \in \mathbb{Z}$. Solving for $x$ yields: $x = \frac{-(3+2y) \pm \sqrt{(3+2y)^2 - 4(1)(y^2+2y+2-K^2)}}{2} = \frac{-(3+2y) \pm \sqrt{4K^2 + 4y + 1}}{2}$ (iii). If we choose $K = y$, then then the discriminant in (iii) is a square, namely $(2y+1)^2$, which yields: $x= \frac{-(3+2y) \pm (2y+1)}{2} \Rightarrow x = -1, -2(y+1)$ which is a contradiction since $x \in \mathbb{N} \cup \{0\} \Rightarrow a \notin \mathbb{Q}$.