How does one show that for $k \in \mathbb{Z_+},3\mid2^{2^k} +5$ and $7\mid2^{2^k} + 3, \forall \space k$ odd. 
For $k \in \mathbb{Z_+},3\mid2^{2^k} +5$ and $7\mid2^{2^k} + 3, \forall \space k$ odd. 

Firstly,
$k \geq 1$
I can see induction is the best idea:
Show for $k=1$:
$2^{2^1} + 5 = 9 , 2^{2^1} + 3 = 7$
Assume for $k = \mu$
so: $3\mid2^{2^\mu} + 5 , \space 7\mid2^{2^\mu} + 3$
Show for $\mu +2$
Now can anyone give me a hint to go from here? My problem is being able to show that $2^{2^{\mu+2}}$ is divisible by 3, I can't think of how to begin how to show this. 

You have already shown that the base cases hold.
Assume $3\mid 2^{2^k}+5$. Then $2^{2^k}\equiv 1$ mod $3$. Hence:
$$2^{2^{k+1}}=2^{2^k*2}=\left(2^{2^k}\right)^2\equiv 1 \text{ mod } 3$$
Hence $3\mid 2^{2^{k+1}}+5$.
In the same way:
Assume $7\mid 2^{2^{k}}+3$. Then $2^{2^{k}}\equiv 4$ mod $7$. Hence:
$$2^{2^{k+2}}=\left(2^{2^k}\right)^4\equiv 4^4 \text{ mod } 7$$
And since $4^4=256=36*7+4$, we see that $256\equiv 4\text{ mod }7$. So $7\mid 2^{2^{k+2}}+3$.
