Just to have the full equation we're solving for reference:$$\sqrt{2x^2+4} = x+k.$$

Short Answer: When dealing with extraneous solutions, understanding what can go WRONG is typically enough. In this case, squaring both sides can introduce extraneous solutions because the correct statement would be$$2x^2+4 = (x+k)^2 \text{ if and only if } \pm \sqrt{2x^2+4} = x+k.$$Note this is saying that (for real solutions) IF there are extraneous solutions (to our original equation) they are actually solutions to $- \sqrt{2x^2+4} = x+k$. Thus if you are convinced that $k\leq -\sqrt{2}$ are the only times that don't work, the other times will work.

If this isn't convincing here, after squaring we can just use the quadratic formula to solve for $x$. This gives $x = k \pm \sqrt{4k^2-8}$. From here it's not too hard (but a little tedious) to directly check that this produces valid solutions for $k\geq \sqrt{2}$.

Alternatively, even a rough picture of the graphs should be very convincing here. $y=x+k$ is a line. The graph of $y=\sqrt{2x^2+4}$ is the square root of a parabola, so it is somewhat of a "flattened out" parabola. From there it should be clear that having real solutions for a large enough value of $k$ will work. You can see the graph on Desmos by clicking here.