Find all positive integers $x, y, z$ and $t$ such that

$$
2^{x} \cdot 3^{y} + 5^{z} = 7^{t}
$$
Reducing modulo 3 we get $5^{z} \equiv 1$, therefore $z$ is even, $z=2 c, c \in \mathbb{N}$.

Next we prove that $t$ is even:

Obviously, $t \geq 2$. Let us suppose that $t$ is odd, say $t=2 d+1, d \in \mathbb{N}$. The equation becomes $2^{x} \cdot 3^{y}+25^{c}=7 \cdot 49^{d}$. If $x \geq 2$, reducing modulo 4 we get $1 \equiv 3$, a contradiction. And if $x=1$, we have $2 \cdot 3^{y}+25^{c}=7 \cdot 49^{d}$ and reducing modulo 24 we obtain

$$
2 \cdot 3^{y}+1 \equiv 7 \Rightarrow 24 \mid 2\left(3^{y}-3\right) \text {, i.e. } 4 \mid 3^{y-1}-1
$$

which means that $y-1$ is even. Then $y=2 b+1, b \in \mathbb{N}$. We obtain $6 \cdot 9^{b}+25^{c}=7 \cdot 49^{d}$, and reducing modulo 5 we get $(-1)^{b} \equiv 2 \cdot(-1)^{d}$, which is false for all $b, d \in \mathbb{N}$. Hence $t$ is even, $t=2 d, d \in \mathbb{N}$, as claimed.

Now the equation can be written as

$$
2^{x} \cdot 3^{y}+25^{d}=49^{d} \Leftrightarrow 2^{x} \cdot 3^{y}=\left(7^{d}-5^{c}\right)\left(7^{d}+5^{c}\right)
$$

As $\operatorname{gcd}\left(7^{d}-5^{c}, 7^{d}+5^{c}\right)=2$ and $7^{d}+5^{c}>2$, there exist exactly three possibilities:
(1) $\left\{\begin{array}{l}7^{\mathrm{d}}-5^{\mathrm{d}}=2^{\mathrm{x-1}} \\ 7^{\mathrm{d}}+5^{\mathrm{d}}=2 \cdot 3^{y}\end{array} ;\right.$
(2) $\left\{\begin{array}{l}7^{\mathrm{d}}-5^{\mathrm{d}}=2 \cdot 3^{\mathrm{y}} \\ 7^{\mathrm{d}}+5^{\mathrm{d}}=2^{\mathrm{x-1}}\end{array}\right.$;
(3) $\left\{\begin{array}{l}7^{d}-5^{d}=2 \\ 7^{d}+5^{d}=2^{x-1} \cdot 3^{y}\end{array}\right.$

## Case (1)

We have $7^{d}=2^{x-2}+3^{y}$ and reducing modulo 3, we get $2^{x-2} \equiv 1(\bmod 3)$, hence $x-2$ is even, i.e. $x=2 a+2, a \in \mathbb{N}$, where $a>0$, since $a=0$ would mean $3^{y}+1=7^{d}$, which is impossible (even $=$ odd).

We obtain

$$
7^{d}-5^{d}=2 \cdot 4^{a} \stackrel{\bmod 4}{\Rightarrow} 7^{d} \equiv 1(\bmod 4) \Rightarrow d=2 e, e \in \mathbb{N}
$$

Then we have

$$
49^{e}-5^{d}=2 \cdot 4^{a} \stackrel{\bmod 8}{\Rightarrow} 5^{c} \equiv 1(\bmod 8) \Rightarrow c=2 f, f \in \mathbb{N}
$$

We obtain $49^{e}-25^{f}=2 \cdot 4^{a} \stackrel{\text { mod } 3}{\Rightarrow} 0 \equiv 2(\bmod 3)$, false. In conclusion, in this case there are no solutions to the equation.

## Case (2)

From $2^{x-1}=7^{d}+5^{c} \geq 12$ we obtain $x \geq 5$. Then $7^{d}+5^{c} \equiv 0(\bmod 4)$, i.e. $3^{d}+1 \equiv 0(\bmod 4)$, hence $d$ is odd. As $7^{d}=5^{c}+2 \cdot 3^{y} \geq 11$, we get $d \geq 2$, hence $d=2 e+1, e \in \mathbb{N}$.

As in the previous case, from $7^{d}=2^{x-2}+3^{y}$ reducing modulo 3 we obtain $x=2 a+2$ with $a \geq 2$ (because $x \geq 5$ ). We get $7^{d}=4^{a}+3^{y}$ i.e. $7 \cdot 49^{e}=4^{a}+3^{y}$, hence, reducing modulo 8 we obtain $7 \equiv 3^{y}$ which is false, because $3^{y}$ is congruent either to 1 (if $y$ is even) or to 3 (if $y$ is odd). In conclusion, in this case there are no solutions to the equation.

## Case (3)

From $7^{d}=5^{c}+2$ it follows that the last digit of $7^{d}$ is 7, hence $d=4 k+1, k \in \mathbb{N}$.

If $c \geq 2$, from $7^{4 k+1}=5^{c}+2$ reducing modulo 25 we obtain $7 \equiv 2(\bmod 25)$ which is false. For $c=1$ we get $d=1$ and the solution $x=3, y=1, z=t=2$.