There are no numbers $x \in [1, \infty)$ with the property in the statement: if $x \ge 1$, then $[x] \ge 1$ and, since $\{x\} \in [0, 1)$, we obtain
$$4 = 3^x + 3^{[x]} + 3^{\{x\}} \ge 3 + 3 + 1 = 7,$$
contradiction.

There are no numbers $x \in (-\infty, -1)$ with the property in the statement: $x < -1$, then $[x] \le -2$ and, since $\{x\} \in [0, 1)$, we obtain
$$4 = 3^x + 3^{[x]} + 3^{\{x\}} < 3^{-1} + 3^{-2} + 3^1 = 3\frac{4}{9},$$
contradiction.

If $x \in [0, 1)$, then $[x] = 0$, $\{x\} = x$, so the given equality becomes
$$3^x + 1 + 3^x = 4 \Leftrightarrow 2 \cdot 3^x = 3.$$
We obtain the solution $x_1 = 1 - \log_3 2 \in [0, 1)$.

If $x \in [-1, 0)$, then $[x] = -1$, $\{x\} = x + 1$, so the given equality becomes
$$3^x + \frac{1}{3} + 3^{x+1} = 4 \Leftrightarrow 4 \cdot 3^x = \frac{11}{3}.$$
We obtain the solution $x_2 = \log_3 \frac{11}{12} \in [-1, 0)$.