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

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 $t$.