Introduce a new sequence $v_n = u_n - 1$. We have $v_0 = 1$ and the given recursion translates into
$$v_n + 1 = \frac{2v_{n-1} + 3}{v_{n-1} + 3}, \text{ i.e. } v_n = \frac{v_{n-1}}{v_{n-1} + 3} = 1 - \frac{3}{v_{n-1} + 3}.$$
We have to prove that $0 < v_n < \frac{2}{3^n}$. As $v_0 = 1 > 0$, the recursion easily implies that $v_n > 0$ for all $n \ge 0$.
Let us now show by induction that $v_n < \frac{2}{3^n}$. The start at $n = 0$ is clear, as $v_0 = 1 < 2$. Assume now that the inequality holds true for $n = k \ge 0$, i.e. $v_k < \frac{2}{3^k}$. This assumption implies
$$v_k + 3 < \frac{2 + 3^{k+1}}{3^k} \quad \text{and so} \quad -\frac{3}{v_k + 3} < -\frac{3^{k+1}}{2 + 3^{k+1}}, \quad \text{hence}$$
$$v_{k+1} = 1 - \frac{3}{v_k + 3} < 1 - \frac{3^{k+1}}{2 + 3^{k+1}} = \frac{2}{2 + 3^{k+1}} < \frac{2}{3^{k+1}}.$$

The sequence $(u_n)_n^\infty$ begins $2, \frac{5}{4}, \frac{14}{13}, \frac{41}{40}, \frac{122}{121}, \frac{365}{364}, \dots$, and helped by the occurrence of $3^n$ in the statement of the question, we check that the sequence $(3^n)_n^\infty$ begins $1, 3, 9, 27, 81, 243, 729, \dots$, and we note the pattern $\left(3^{n+1} \pm 1\right)/2$ in the form of the numerators and denominators of the terms $u_n$. We therefore conjecture
$$u_n = \frac{3^{n+1} + 1}{3^{n+1} - 1}, \text{ for all } n \ge 0.$$
We prove the conjecture by simple induction. The formula is trivial for $n = 0$. Assuming the validity of the conjecture for $n = k \ge 0$, we have
$$u_{k+1} = \frac{2u_k + 1}{u_k + 2} = \frac{2\left(\frac{3^{k+1}+1}{3^{k+1}-1}\right) + 1}{\left(\frac{3^{k+1}+1}{3^{k+1}-1}\right) + 2} = \frac{2(3^{k+1} + 1) + (3^{k+1} - 1)}{(3^{k+1} + 1) + 2(3^{k+1} - 1)} = \frac{3^{k+2} + 1}{3^{k+2} - 1},$$
as asserted. This proves the conjecture. It now follows immediately that $u_n > 1$ and that
$$u_n - 1 = \frac{2}{3^{n+1} - 1} < \frac{2}{3^n}.$$