Solutions of $a^{2} - 2b^{2} \equiv 0$ mod $p$

I came across this question in attempting to find $p$ for which $\mathbb{Z}_{p}[\sqrt{2}]$ is a field.

Consider the equation:

$$a^2 - 2b^2 \equiv 0 \enspace \text{mod p}$$

For which primes $p$ is it possible for this equation to have solutions in $a$ and $b$?

I wrote a quick brute force number crunching program to find that for primes 3, 5, 11, 13, and 19 there are no solutions in $a$ and $b$.

However, for primes 7 and 17, there exist solutions.

Is there a discernible condition on $p$ such that this congruence has solutions?

First of all, note that $$a\equiv b\equiv0\pmod p$$ is a solution for any $$p$$; I assume you mean to exclude these trivial solutions. That means we can assume that neither $$a$$ nor $$b$$ is $$0\pmod p$$ (since if one is, the other is too).
In particular, $$b$$ is invertible modulo $$p$$, and so the congruence is equivalent to $$(ab^{-1})^2 \equiv 2\pmod p$$. Therefore if the congruence has nontrivial solutions, so does the congruence $$x^2\equiv2\pmod p$$. Conversely, if $$x^2\equiv2\pmod p$$ has solutions, then take $$a=x$$ and $$b=1$$.
So we've reduced the problem to deciding which primes $$p$$ have "square roots of $$2$$"; in other words, we want to know for which primes $$2$$ is a quadratic residue. And this is a classical theorem: they are exactly the primes congruent to either $$1$$ or $$7$$ modulo $$8$$.