Since $p,q$ are rationals, and assuming $pq\ne 0$, we can always set $q=(p+1)\cdot\frac{b}{a},$ where $a,b$ are integers. Substituting this expression for $q$ into $p^2-xq^2=1$ we get the desired result.

Nice problem. Here's my idea. What $p^2-xq^2=1$ is? Of course, it is a hyperbola. Lemma 1 Let points $X,Y$ have rational coordinates. Then equation of line through $X$ and $Y$ can be expressed as linear equation with rational coefficients ($ax+by=c; a,b,c \in \mathbb{Q}$). Lemma 2 Let line $ax+by=c; a,b,c \in \mathbb{Q}$ pass through point $P(x,y)$. If one of its coordinates ($x$ or $y$) is rational then the second one is rational too. These lemmas are obvious so I left them without proof. $(p,q)=(1,0)$ is a rational point on hyperbola. We need to find all rational points on one. Pick any rational point $(p,q)$ on hyperbola (I'll work only with $x>0$ part the rest is easy obtained by symmetry) and let line through $(1,0)$ and $(p,q)$ intersects ordinate at $(0,c)$. Due to lemmas mentioned above $c \in \mathbb{Q}$. Thus we have bijection between rational numbers and rational solutions of $p^2-xq^2=1$. More precisely: Consider line through $(0,c)$ and $(1,0)$ $q=c-cp$ so $p^2-x(c-cp)^2=1$ $p^2(1-c^2x)+2c^2xp-c^2x-1=0$ It is a quadratic equation wrt $p$ It has two solutions: $p=1$ and $p=\frac{1+c^2x}{c^2x-1}$ Since $c \in \mathbb{Q}$ there exist $a,b \in \mathbb{Z}$ such that $c=\frac{a}{b}$. The rest is trivial.

Some blah. I hadn’t been interested in Israel contests before, but their performance on RMM 2019 changed my mind... Unfortunately there are almost no Israel problems on aops... can someone fix this?