II. (40 points) Find all non-negative integer solutions $(a, b, c)$ to the equation $2^{a}=2^{b}+20^{c}+2020$.

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Let $(a, b, c)$ be a solution.
When $c=0$, we have $2^{a}-2^{b}=2$, hence $a=2, b=1$.
When $c=1$, we have $2^{a}-2^{b}=2040$, hence $a=11, b=3$.
Now consider the case when $c \geqslant 2$.
From $2^{a} \equiv 2^{b}(\bmod 5) \Rightarrow a \equiv b(\bmod 4)$.
Since $a>b$, let $a=b+4 k\left(k \in \mathbf{Z}_{+}\right)$, substituting into the original equation and rearranging gives
$$
2^{b}\left(2^{4 k}-1\right)=2^{2 c+1} \times \frac{5^{c}+505^{c}}{2} .
$$

Since $2^{4 k}-1$ and $\frac{5^{c}+505^{c}}{2}$ are both odd, we have
$$
b=2 c+1 \text {. }
$$

Then $16^{k}-1=\frac{5^{c}+505^{c}}{2}$.
Notice that $16 \equiv 1(\bmod 5)$.
By the lifting the exponent lemma, we know
$$
\begin{array}{l}
v_{5}\left(16^{k}-1\right)=v_{5}(16-1)+v_{5}(k) \\
=1+v_{5}(k),
\end{array}
$$

where $v_{5}(N)$ denotes the exponent of the prime factor 5 in the positive integer $N$.
Thus, from equation (1) we have
$$
\begin{array}{l}
1+v_{5}(k)=v_{5}\left(\frac{5^{c}\left(1+101^{c}\right)}{2}\right) \\
=c+v_{5}\left(\frac{1+101^{c}}{2}\right)=c .
\end{array}
$$

Hence $5^{c-1} \mid k$
$$
\begin{aligned}
\Rightarrow & 16^{5 c-1} \leqslant 16^{k}=1+\frac{5^{c}+505^{c}}{2} \\
& \frac{9}{4} c$, which is a contradiction.
In conclusion, the non-negative integer solutions to the original equation are
$$
(a, b, c)=(2,1,0),(11,3,1) .
$$