Showing that there exists a positive integer $t$ such that $5^t\equiv -3\pmod {2^{n+4}}$ When I deal with number theory, I encounter a problem that seems to be easy but I can't prove.

let $n$ be a positive integer, there exists a positive integer $t$ such that 
  $$5^t\equiv -3\pmod {2^{n+4}}$$

 A: Recall that for $n\geq 2$, the element $5$ has order $2^{n-2}$ in $(\Bbb Z/2^n\Bbb Z)^\times$.
This follows by proving that
$$5^{2^{n-2}}=1+2^nk_n$$
with $2\nmid k_n$.
For arguing by induction on $n$, we have
\begin{align}
5^{2^{n-1}}
&=(1+2^nk_n)^2\\
&=1+2^{n+1}k_n+2^{2n}k_n^2\\
&=1+2^{n+1}k_n(1+2^{n-1}k_n)
\end{align}
thus $2\nmid k_{n+1}=k_n(1+2^{n-1}k_n)$.
Consequently, we have a surjective groups homomorphism
$\sigma_n:(\Bbb Z/2^n\Bbb Z)^\times\to\{\pm 1\}$
such that $\operatorname{Ker}\sigma_n=\langle 5\rangle$.
The canonical ring homomorphism $\varepsilon_n:\Bbb Z/2^n\Bbb Z\to\Bbb Z/4\Bbb Z$ induces a surjective group homomorphism $\varepsilon_n^\times:(\Bbb Z/2^n\Bbb Z)^\times\to(\Bbb Z/4\Bbb Z)^\times$ and we have a commutative diagram:
$\require{AMScd}$
\begin{CD}
(\Bbb Z/2^n\Bbb Z)^\times@>\varepsilon_n^\times>>(\Bbb Z/4\Bbb Z)^\times\\
@V\sigma_nVV@V\sim V\sigma_2V\\
\{\pm 1\}@=\{\pm 1\}
\end{CD}
From this follows that $x\in\operatorname{Ker}\sigma_n$ if and only if $x\equiv 1\pmod 4$.
In particular, $-3\equiv 1\pmod 4$, hence $-3\in\operatorname{Ker}\sigma_n=\langle 5\rangle$, that's $-3\equiv 5^t\pmod{2^n}$ for some $t$.
A: Claim 1. For any $n\ge0$, $5^{2^n}-1=2^{n+2}.odd$.
By induction, $$5^{2^{n+1}}-1=(5^{2^n}-1)(5^{2^n}+1)=2^{n+2}.odd.2.odd=2^{n+3}.odd$$ since in general $5^m+1=1^m+1=2\pmod{4}$. Of course, $5^1-1=2^2.1$.
Claim 2. For every $k$, there is an $r$ such that for any $n$, $2^k|(5^{2^{k-2}n+r}+3)$.
For $k=2$, $5^n+3=1+3=0\pmod{4}$.
Suppose $2^k|(5^{2^{k-2}n+r}+3)$; then either $2^{k+1}|(5^{2^{k-1}n+r}+3)$ satisfies the claim, or else $5^{2^{k-1}n+r}+3=2^k.odd$. In this case, \begin{align*} 5^{2^{k-2}(2n+1)+r}+3&=5^{2^{k-1}n+r}.5^{2^{k-2}}+3\\
&=(5^{2^{k-1}n+r}+3)5^{2^{k-2}}-3(5^{2^{k-2}}-1)\\
&=2^k.odd.5^{2^{k-2}}-3.2^k.odd\\
&=2^k(odd-odd)\\
&=2^{k+1}a\end{align*}
Hence for every $k\ge2$ there are an infinite number of $t$ such that $5^t=-3\pmod{2^k}$.
