All equilibria found with backwards induction on a tree of a perfect information game are Nash equilibria, but in general the reverse is not true:
y
(1)---+---(0, 20) ← Also a Nash equilibrium when (2) announces
| n y he will play y: (0, 20) > (-10, -2)
+---(2)---+---(-10, -2)
| n
+---( +5, -1) ← Solution through backwards induction
(In class, we've called this the "icecream game", (1) is the mom who needs to decide whether to buy his son an ice cream and (2) is the son who needs to decide whether to cry or not about it.)
However, Chess (or Checkers, or Tic Tac Toe) are different from the "icecream game" because the payoffs are either (1, −1) (white win), (−1, 1) (black win) or (0,0) (draw).
Do these games still allow Nash equilibria that can't be found through backward induction?
Chess is a zero-sum game, so all Nash equilibria will lead to the same of three outcomes: White wins. Black wins. Draw. By Zermelo's theorem, in the first case white can force a win. In the second case, black can force a win. In the third case, both can force a draw.
It is not known which case holds, but in the first two cases, there will be many Nash equilibria (a whole continuum). If one player plays a strategy that guarantees a win for her, her strategy combined with any strategy of the other player together will constitute a Nash equilibrium. Even if both players can force a draw, there may be several ways to do so.
So in conclusion, chess has probably more than one Nash equilibrium. If there is only one Nash equilibrium, it will end in a draw.
Remark: Zermelo did not use backward induction and his proof was not based on chess having a stopping rule.