$\{(x, y, z)\} = \{(0, 1, 1), (1, 1, 2), (0, 3, 2), (2, 5, 4)\}$.
We rewrite the equation in the form $7^x + 2^y = 3^z$. Note that for $y = 0$ the left-hand side of the equation is an even integer while the right-hand side is an odd integer. So, $y > 0$.
Consider three cases: $y = 1$, $y = 2$ and $y \ge 3$.
1) $y = 1$. In this case we have
$$7^x + 2 = 3^z. \quad (1)$$
If $x = 0$, then $z = 1$; if $x = 1$, then $z = 2$. Thus, it remains to consider the case $x > 2$, $z > 2$. In this case $(7^x + 2) \ge 3^3$. Consider the residues of $7^n$ modulo 27:
$$7 \rightarrow -5 \rightarrow -8 \rightarrow -2 \rightarrow 13 \rightarrow 10 \rightarrow -11 \rightarrow 4 \rightarrow 1.$$
It follows that $x \equiv 4 \pmod 9$. Note that $7^6 + 7^3 + 1 = 117993 = 3 \cdot 37 \cdot 1063$, so $(7^9 - 1) \equiv (7^6 + 7^3 + 1) \equiv 37$. Thus, $7^9 \equiv 1 \pmod{37}$ and then
$$7^x \equiv 7^4 \equiv 49^2 \equiv 12^2 \equiv 144 \equiv 33 \pmod{37}.$$
Hence $7^x + 2 \equiv 35 \pmod{37}$.
On the other hand, considering the residues of $3^n$ modulo 37, we have
$$3 \rightarrow 9 \rightarrow 27 \rightarrow 7 \rightarrow 21 \rightarrow 26 \rightarrow 4 \rightarrow 12 \rightarrow 36 \rightarrow 34 \rightarrow 28 \rightarrow 10 \rightarrow 30 \rightarrow 16 \rightarrow 11 \rightarrow 33 \rightarrow 25 \rightarrow 1.$$
We see that $3^z \not\equiv 35 \pmod{35}$ for all $z$. Therefore, there are no solutions of (1) for $x > 2$, $z > 2$.
2) $y = 2$. We have $3^z = 7^x + 4 \equiv 2 \pmod 3$, a contradiction.
3) $y \ge 3$. Then $7^x \equiv 3^z \pmod 8$, whence $x \equiv 2 \pmod 2$ and $z \equiv 2 \pmod 2$. Let $x = 2x_1$, $z = 2z_1$. Now the initial equation can be rewritten in the form
$$(3^{z_1} - 7^{x_1})(3^{z_1} + 7^{x_1}) = 2^y.$$
So, it follows that
$$3^{z_1} - 7^{x_1} = 2^a, \quad (2)$$
$$3^{z_1} + 7^{x_1} = 2^b, \quad (3)$$
and, since $(3^{z_1} - 7^{x_1}) \equiv 2$, we have $b > a \ge 1$. Summing (2) and (3), we obtain $2^a + 2^b = 2 \cdot 3^{z_1}$. Hence $a = 1$. Then (2) can be rewritten as $3^{z_1} - 7^{x_1} = 2$, or $3^{z_1} = 7^{x_1} + 2$, which coincides with (1). Therefore, we have $\{(x_1, z_1)\} = \{(0, 1), (1, 2)\}$, and $b$ is equal to 2 and 4 respectively.
Finally, in this case we have $\{(x, y, z)\} = \{(0, 3, 2), (2, 5, 4)\}$.

Answer: $\{(x, y, z)\} = \{(0, 1, 1), (1, 1, 2), (0, 3, 2), (2, 5, 4)\}$.

We rewrite the equation in the form $7^x + 2^y = 3^z$. Note that for $y = 0$ the left-hand side of the equation is an even integer while the right-hand side is an odd integer. So, $y > 0$.

Consider three cases: $y = 1$, $y = 2$ and $y \ge 3$.

1)
$y = 1$. In this case we have
$$7^x + 2 = 3^z. \tag{1}$$
If $x = 0$, then $z = 1$; if $x = 1$, then $z = 2$. Thus, it remains to consider the case $x > 2$, $z > 2$. In this case $(7^x + 2) \div 3^3$. Consider the residues of $7^n$ modulo $27$:
$$7 \rightarrow -5 \rightarrow -8 \rightarrow -2 \rightarrow 13 \rightarrow 10 \rightarrow -11 \rightarrow 4 \rightarrow 1.$$
It follows that $x \equiv 4 \pmod{9}$. Note that $7^6 + 7^3 + 1 = 117993 = 3 \cdot 37 \cdot 1063$, so $(7^9 - 1) \div (7^6 + 7^3 + 1) \div 37$. Thus, $7^9 \equiv 1 \pmod{37}$ and then
$$7^x \equiv 7^4 \equiv 49^2 \equiv 12^2 \equiv 144 \equiv 33 \pmod{37}.$$
Hence $7^x + 2 \equiv 35 \pmod{37}$.

On the other hand, considering the residues of $3^n$ modulo $37$, we have
$$3 \rightarrow 9 \rightarrow 27 \rightarrow 7 \rightarrow 21 \rightarrow 26 \rightarrow 4 \rightarrow 12 \rightarrow 36 \rightarrow 34 \rightarrow 28 \rightarrow 10 \rightarrow 30 \rightarrow \\ \rightarrow 16 \rightarrow 11 \rightarrow 33 \rightarrow 25 \rightarrow 1.$$
We see that $3^z \not\equiv 35 \pmod{37}$ for all $z$. Therefore, there are no solutions of (1) for $x > 2$, $z > 2$.

2) $y = 2$. We have $3^z = 7^x + 4 \equiv 2 \pmod{3}$, a contradiction.

3) $y \ge 3$. Then $7^x \equiv 3^z \pmod{8}$, whence $x \div 2$ and $z \div 2$. Let $x = 2x_1$, $z = 2z_1$. Now the initial equation can be rewritten in the form
$$(3^{z_1} - 7^{x_1})(3^{z_1} + 7^{x_1}) = 2^y.$$
So, it follows that
$$3^{z_1} - 7^{x_1} = 2^a, \qquad (2)$$
$$3^{z_1} + 7^{x_1} = 2^b, \qquad (3)$$
and, since $(3^{z_1} - 7^{x_1}) \div 2$, we have $b > a \ge 1$. Summing (2) and (3), we obtain $2^a + 2^b = 2 \cdot 3^{z_1}$. Hence $a = 1$. Then (2) can be rewritten as $3^{z_1} - 7^{x_1} = 2$, or $3^{z_1} = 7^{x_1} + 2$, which coincides with (1). Therefore, we have $\{(x_1, z_1)\} = \{(0, 1), (1, 2)\}$, and $b$ is equal to $2$ and $4$ respectively.

Finally, in this case we have $\{(x, y, z)\} = \{(0, 3, 2), (2, 5, 4)\}$.