You started out well enough. You got to: the assumption of a solution implies $(p|q) = 1$ and so $(q|p ) = -1.$
Next, we think we have $x^2 - q y^2 = p,$ in particular
$$ x^2 \equiv q y^2 \pmod p. $$ If $y \neq 0 \pmod p,$ then $y$ has a multiplicative inverse $\pmod p,$ and
$$ \left( \frac{x}{y} \right)^2 \equiv q \pmod p. $$ Well this contradicts $(q|p ) = -1,$ so in fact $y$ is divisible by $p.$ But then $x^2 - q y^2 = p$ says that $x$ is also divisible by $p.$ Thus $$ x^2 - q y^2 \equiv 0 \pmod {p^2}, $$ which contradicts $x^2 - q y^2 = p.$
An answer was posted using just $\pmod 4,$ which is worth knowing.
However, the part about a value of the Legendre symbol implying that $x,y$ are both divisible by some prime is a standard ingredient in quadratic forms and will come up again and again.
The full version is this: if I have a quadratic form $$ f(x,y) = a x^2 + b x y + c y^2, $$ it has a "discriminant" that is the same as the term under the square root in the Quadratic Formula,
$$ \Delta = b^2 - 4 a c. $$
Now, $\Delta$ is allowed to be positive or negative, however if it is positive we do not permit it to be a perfect square. The technical term for binary quadratic forms with square discriminant is "stupid."
Now, suppose we have some prime $p$ that does not divide $\Delta,$ and for which
$$ (\Delta | p) = -1. $$ Note that we allow $p \equiv \pm 1 \pmod 4$ here.
What happens if
$$ a x^2 + b x y + c y^2 \equiv 0 \pmod p? $$
Just to make it easy on myself, demand that $a$ not be divisible by $p.$
If so, it is still true if I multiply both sides by $4a,$ I get
$$ 4 a^2 x^2 + 4 a b x y + 4 ac y^2 \equiv 0 \pmod p, $$ and
$$ 4 a^2 x^2 + 4 a b x y + b^2 y^2 - b^2 y^2 + 4 ac y^2 \equiv 0 \pmod p, $$
$$ (2 a x + b y)^2 - \Delta y^2 \equiv 0 \pmod p, $$ so
$$ (2 a x + b y)^2 \equiv \Delta y^2 \pmod p. $$
If we assume that $y$ is not divisible by $p,$ it has a multiplicative inverse $\pmod p,$ and we get a square equivalent to $\Delta \pmod p,$ not permitted. So actually $y$ is divisible by $p.$ Next $(2ax+ by)$is divisible by $p,$ and finally $x$ is as well as $y$ because I demanded that $a$ n not be. So there.
The bit about $a$ not being divisible by $p$ is not critical. As $\Delta$ is not divisible by $p$ it is impossible for $a,b,c$ to all be divisible by $p.$ And hilarity ensues. It has come back to me, as in a dream. If $a$ is divisible by $p,$ simply switch the argument to $c,$ multiply by $4c$ rather than $4a$ and so on. We might worry about both $a,c$ being divisible by $p,$ but in that case
$\Delta \equiv b^2 \pmod p $ and $(\Delta | p) = (b^2 | p) =1, $ contradicting $(\Delta | p) = -1.$