I will prove that there exists infinitely many numbers $n^4$ such that $n^4=x^2+2y^2$ in which $x$ is odd and $y$ is even and $gcd(x,y)=1$ . Clearly we have that $n^4$ is a good number. Reason : $x^2$ has even prime factors and since $y^2$ has even prime factors $\implies$ $2y^2$ has odd prime factors. The goal is to construct a good number from previous good numbers Note that we have : $3^4=7^2+2*4^2$ By lagrange formula we have that $(a^2+mb^2)(c^2+md^2)=(ac-mbd)^2+m(bc+ad)^2$ By putting $c=a$ and $d=b$ and $m=2$ we get that $(a^2+2b^2)(a^2+2b^2)=(a^2-2b^2)^2+2(2ab)^2$ Assume that we have $(a^2+2b^2)=n_1^4$ , and $n_1^4$ is the largest number we know of the form $a^4$ and good number . Therefor by the formula above we can see that $n_1^8=(a^2-2b^2)+2(2ab)^2 \implies n_1^8$ is a good number , so we have found a bigger number of the form $a^4$ such that it is a good number . ( Note that $a^2-2b^2$ and $2ab$ are coprime) So we start form $3^4$ and construct $9^4$ and $81^4$ $\dots$ $Q.E.D$

I may be missing something but doesn't $(2k)^4$ work? $16k^4=14k^4+2k^4$ ???????

Taha1381 wrote: I may be missing something but doesn't $(2k)^4$ work? $16k^4=14k^4+2k^4$ ??????? I believe that $14k^4$ and $2k^4$ are not coprime.