Is everything perfect?

Note that $2+2\sqrt{28n^2+1}=m\longrightarrow 4(28n^2+1)=m^2-4m+4\longrightarrow m=2k\longrightarrow 28n^2+1=k^2-2k+1$

From that: $28n^2=k^2-2k\longrightarrow k=2q\longrightarrow 28n^2=4q^2-4q\longrightarrow 7n^2=q(q-1)$

Here $\gcd{(q, q-1)}=1$ . There are two cases:

$(i)$ $q=7x^2, q-1=y^2\longrightarrow 7x^2-y^2=1$ . Since $y^2\not\equiv -1\pmod 7$ , this case doesn't have a solution.

$ii$ $q=x^2, q-1=7y^2$ In this case $m=2k=4q=4x^2=(2x)^2$ .

Hence it is always a perfect square.

It is suffices to show that there are conterexamples for the other statements. Let $n=24$ . Then $m=256=16^2$ , and $m$ is not a $7^{\text{th}}$ power or a perfect cube. One can find a counterxample for the fourth power.