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For any natural number $x$, $x^2 + 5$ is not divisible by $4$

Or is proof by cases the only way to go about proving this?

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    It can be done by induction, most easily by starting with cases $1$ and $2$ (neither of whose squares plus $5$ are divisible by $4$), and then proceeding by induction in steps of $2$. (That is to say, $(x+2)^2+5 = x^2+4x+4+5 = (x^2+5)+4(x+1)$.) – 2017-02-09
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    Do you know how to proceed after the comment of Brian, i.e. is your question answered now? – 2017-02-09

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Besides induction and "by cases" maybe consider a direct proof, too. If $\;4 \mid x^2 +5\,$ then $x^2$ must be odd, so $x$ itself must be odd. But then $x=2y+1\,$ so $x^2+5=4y^2+4y+6 \equiv 2 \pmod 4\,$.