prove that for every integer,$ n$, greater than $1$,$ (3^{2n+1}) + (5^{2n-1})$ is divisible by $16$ Could someone please help me out with this proof?

Prove that for every integer $n≥1$, $3^{2n+1} + 5^{2n-1}$, is divisible by $16$.

I get to a point where I have...

$$3^{2k+1} \cdot 3^2 + 5^{2k-1} \cdot 5^2 $$

Thanks!
The order of $3$ and $5$ in $(\mathbb{Z}/16\mathbb{Z})^*$ is four, hence you need to check the claim only for two cases: $n=1$ and $n=2$. For the remaining cases, just use the fact that:
$$3^{2(n+2)+1}\equiv 3^{2n+1}\pmod{16},\qquad 5^{2(n+2)-1}\equiv 5^{2n-1}\pmod{16}.$$
I generalized this here:
When is there an $m$ that divides $u^{an+b}+v^{cn+d}$ for all $n$

I asked:
Find conditions on
 positive integers
$u, a, v, c$
and non-negative integers
$b, d$,
such that
there exists an $m$
which divides
$f(n) = u^{an+b}+v^{cn+d}$
for all $n$.
Determine $m$
in terms of
$u, a, b, v, c, d$.

One conclusion was
that if
$(u^a-v^c)|(u^b+v^d)$,
it divides $f(n)$
for all $n$.

In this case,
putting $n+1$ for $n$
(so $d$ is not negative),
$(u, a, v, c, b, d)
=
(3, 2, 5, 2, 3, 1)
$,
so
$u^a-v^c
=3^2-5^2
=-16
$
and
$u^b+v^d
=3^3+5^1
=27+5
=32
$.
Since $-16$ divides $32$,
$16$ divide all terms.