Solution:
$1997^{a} + 15^{b} = 2012^{c} \Rightarrow 1 + (-1)^{b} \equiv 0 \pmod{4}$, so $b$ is an odd number.
$1997^{a} + 15^{b} = 2012^{c} \Rightarrow 1 + 0 \equiv 2^{c} \pmod{3}$, so $c$ is even, say $c = 2c_{1}$.
We intend to consider the given equation modulo $8$ and for this reason we discern two cases:

(1): $c = 1$. Clearly then $a = b = 1$ and $a = b = c = 1$ is a solution. This is actually the only solution of the given equation since in the remaining case where $c > 1$ it will be shown that there exist no solution.

(2): $c > 1$. Then $2012^{c} = (4 \cdot 503)^{c}$ is a multiple of $8$ and
$1997^{a} + 15^{b} = 2012^{c} \Rightarrow 5^{a} + (-1)^{b} \equiv 5^{a} + (-1) \equiv 0 \pmod{8}$, so $a$ is even, say $a = 2a_{1}$. Hence
$$3^{b} \cdot 5^{b} = 15^{b} = 2012^{c} - 1997^{a} = \left(2012^{c_{1}} - 1997^{a_{1}}\right) \cdot \left(2012^{c_{1}} + 1997^{a_{1}}\right)$$
Observe that $2012^{c_{1}} - 1997^{a_{1}}$, $2012^{c_{1}} + 1997^{a_{1}}$ are both greater than $1$ and prime to each other as $\gcd\left(2012^{c_{1}} - 1997^{a_{1}}, 2012^{c_{1}} + 1997^{a_{1}}\right) = \gcd\left(2012^{c_{1}} - 1997^{a_{1}}, 2 \cdot 1997^{a_{1}}\right) = 1$. So there exist two cases:
$$\text{Case 1:} \begin{aligned} & 2012^{c_{1}} - 1997^{a_{1}} = 5^{b} \\ & 2012^{c_{1}} + 1997^{a_{1}} = 3^{b} \end{aligned}, \quad \text{Case 2:} \begin{aligned} & 2012^{c_{1}} - 1997^{a_{1}} = 3^{b} \\ & 2012^{c_{1}} + 1997^{a_{1}} = 5^{b} \end{aligned}$$

Case 1:
$$\begin{gathered} 2012^{c_{1}} - 1997^{a_{1}} = 5^{b} \Rightarrow 2^{c_{1}} - 2^{a_{1}} \equiv 0 \pmod{5} \Rightarrow c_{1} \equiv a_{1} \pmod{5} \\ 2012^{c_{1}} + 1997^{a_{1}} = 3^{b} \Rightarrow 2^{c_{1}} + 2^{a_{1}} \equiv 0 \pmod{5} \Rightarrow c_{1} \equiv a_{1} + 1 \pmod{5} \end{gathered}$$
a contradiction.

Case 2:
$$\begin{aligned} & 2012^{c_{1}} - 1997^{a_{1}} = 3^{b} \\ & 2012^{c_{1}} + 1997^{a_{1}} = 5^{b} \end{aligned}$$
Since $b$ is an odd number we get $2012^{c_{1}} + 1997^{a_{1}} \equiv 5^{b} \pmod{3} \Rightarrow 2^{c_{1}} + 2^{a_{1}} \equiv 5^{b} \equiv 2 \pmod{3}$ so $a_{1}, c_{1}$ are even numbers, say $a_{1} = 2a_{2}, c_{1} = 2c_{2}$. Then
$$\left(2012^{c_{2}} - 1997^{a_{2}}\right) \cdot \left(2012^{c_{2}} + 1997^{a_{2}}\right) = 3^{b}$$
But $\gcd\left(2012^{c_{2}} - 1997^{a_{2}}, 2012^{c_{2}} + 1997^{a_{2}}\right) = 1$ and the above implies $2012^{c_{2}} - 1997^{a_{2}} = 1$. But then mod $4$, we get $0 - 1 \equiv 3 \pmod{4}$, a contradiction.

Therefore there exists no solution for $c > 1$.

Hence $a = b = c = 1$ is the only solution.