How many primes $p$ are there such that $2p^{3} + 206$ is a perfect square?
My approach: Let the square be $k^{2}$, then
$$2p^{3} + 206 = k^{2}$$ $$2p^{3}=k^{2} - 206$$
$$2p^{3}=(k+√206)(k-√206)$$
Now, let $$(k+√206)=p^{3}; (k-√206)=2$$ and
$$(k-√206)=p^{3}; (k+√206)=2$$
But I couldn't solve it further. Are there no such primes? Am I on the right track? Please help.
I got $19$ by hit-and-trial. Is there any analytic way?