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I have seen problems in the past along these lines:

If integers $a$ and $b$ are prime and if $a^2+b^2+2017=m^2$, then find the two prime numbers.

I have no idea about how to approach this kind of problem.

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    Note that square numbers can only be $0$ or $1\pmod 4$ – 2017-01-19

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Not sure if there are any good general methods, I think it comes down to finding out what you can in each specific case.

Here you could note that if $a,b$ are odd, then $a^2$ and $b^2$ have remainder $1$ when divided by $4$; so the LHS has remainder $3$ when divided by $4$; but the RHS cannot have remainder $3$. So this is impossible.

If one of $a,b$ is odd and the other is even, then LHS has remainder $2$ when divided by $4$; but the RHS cannot have remainder $2$. So this is also impossible.

Therefore the only possibility is $a=b=2$. We then check that $a^2+b^2+2017$ is indeed a square, namely, $45^2$.