Find all natural numbers $n$ so that $4^{m}+ 2^{n}+ 29$ cannot be a perfect square for any natural number $m$
I try to rewrite the $4^{m}+ 2^{n}+ 29$ in the form of a square but I can' t. I know the Fermat Theorem, although it can' t help me to solve this problem.
Turning some comments into a complete answer, it was noted that in order for $4^m+2^n+29$ to be a square, $n$ must be even, since for $n$ odd we have $4^m+2^n+29\equiv2$ mod $3$. Writing $n=2k$, the expression can be written as $4^m+4^k+29$, so we may as well restrict to $m\le k$ (but remember that we've done so).
Now $29\equiv5$ mod $8$, so $4^m+4^k+29$ cannot be a square if $1\lt m\le k$, since odd squares are congruent to $1$ mod $8$. So we only need to consider the cases $m=0$ and $m=1$.
If $m=0$ we have $4^k+30$, which cannot be a square if $k\ge1$, since $30\equiv2$ mod $4$. Nor is $4^0+30=31$ a square.
If $m=1$, we need to see if $4^k+33$ is ever a square. If it is, then we have
$$33=N^2-4^k=N^2-2^{2k}=(N-2^k)(N+2^k)$$
Since $33$ has only two factorizations, $1\times33$ and $3\times11$, we get only two solutions: $(N,k)=(17,4)$ and $(7,3)$. That is, we have $4^1+4^4+29=4^4+4^1+29=289$ and $4^1+4^2+29=4^2+4^1=49$ as the only squares for $4^m+4^k+29$. Reverting to the $2^n$ format, these are
$$4^1+2^8+29=4^4+2^2+29=289$$ and $$4^1+2^4+29=4^2+2^2+29=49$$
Thus $n=2$, $4$, and $8$ are the only exponents for which $4^m+2^n+29$ is ever a perfect square.