2017/023) Find natural number n such that $n^4+33$ is a a perfect square

we need to have $n^4+33 >= (n^2+1)^2$ as next of $x^2$ is $(x+1)^2$
so $33 >= 2n^2 +1$ or $n^2 <=16$ or $n<=4$
n cannot be odd as $n^4+33$ shall be 2 mod 4 and cannot be perfect square
so n = 2 or 4
both are solutions as $2^4+33= 49 = 7^2$ and $4^4+33=289=17^2$