I know this is an over-kill.rewrite the equation as $7^x +1=2.5^y$.For $x \ge 3$ and $y \ge 3$ by zsigmondy's theorem $7^x +1$ has a prime factor $p$ that does not divide $7^2+1=50$ so $p \neq 2,5$.So we get the only two solutions $x=0,2$ and $y=0,2$.so $(x,y)=(0,0),(2,2)$

For $x,y > 2$ write it as $7^2(7^{x-2}-1) = 2\cdot 5^2(5^{y-2} - 1)$. Since the multiplicative order of $7$ modulo $5^2$ is $4$, it follows $4\mid x-2$. Since the multiplicative order of $5$ modulo $7^2$ is $42$, it follows $42\mid y-2$, and $31\mid 5^{42}-1\mid 5^{y-2} - 1$. Thus we need $31\mid 7^{x-2}-1$. Since the multiplicative order of $7$ modulo $31$ is $15$, it follows $15\mid x-2$, so $60\mid x-2$. But then $5^3\mid 7^{x-2}-1$, contradiction.

Good solutions... Can we use zsygmondy's theorem without proof nowadays at IMO or National contests ?

Rewrite the equation as $ 7^x+1=2 \cdot 5^y $. For $y=0$ we get $x=0$. If $y \ge 1$, then the last digit of $7^x$ must be 9, so $x=2a$, where $a$ is an odd positive integer. It follows that $49^a+1=2 \cdot 5^y$ and $y \ge 2$. Using LTE (check the conditions!), we have that $y=v_5(2 \cdot 5^y)=v_5(49^a+1)=v_5(49+1)+v_5(a)$, so $v_5(a)=y-2$, which leads to $a \ge 5^{y-2}$. Denote $5^{y-2}=t, t \ge1$. We have then $ 50t= 50 \cdot 5^{y-2}= 2 \cdot 5^y =49^a+1 \ge 49^{5^{y-2}}+1 = 49^t+1$. But $49^s+1 \ge 50s$ for each $s \ge 2$, so $t=1$, which leads to $y=2$ and $x=2$.

Only with a little knowledge about orders 7^x + 1 = 2*5^y Assume x > 2 --> y > 2. Considering the equation mod 3, we deduce that y is even. Considering the equation mod 4 together with our above result, we deduce that x is even. Considering the equation mod 5 together with our above results, we deduce that x = 4a + 2. Considering the equation mod 7 together with our above results, we deduce that y = 6b + 2. Considering the equation mod 9 together with our above results, we deduce that x = 12a + 2. Considering the equation mod 13 together with our above results, we deduce that y = 12b + 2. Considering the equation mod 49 together with our above results, we deduce that y = 84b + 2. Considering the equation mod 125 together with our above results, we deduce that x = 300a + 110. Considering the equation mod 343 together with our above results, we deduce that y = 4116b + 2102. Since phi(31) = 30 | 300a, then x = 20 (mod phi(31)) --> 7^x + 1 = 5 (mod 31) --> 5^y = 18 (mod 31), which is impossible. Readers are motivated to examine the cases when at least one of x, y is not greater than 2.

rightways wrote: Can we use zsygmondy's theorem without proof nowadays at IMO or National contests ? I think yes!

gobathegreat wrote: Find all nonnegative integer numbers such that $7^x- 2 \cdot 5^y = -1$ Nice solutions , try to solve it with another way. For $(x,y)=(0,0)$ we have the solution If $x,y>0$ Looking $mod3,mod4$ you see that $a,b=0mod2$ set $x=2a$ , $ y=2b$ to transform the equation as $(5^{2b}-1)^2+7^{2a}=5^{4b}$ $(5^{2b}-1,7^{a},5^{2b})=1$ ( so there are naturals $u,v$ with $(u,v)=1$ not together odd and $5^{2b}-1=2uv$ $7^a=u^2-v^2$ $5^{2b}=u^2+v^2$ It easy to see that $u=v-1$ so $5^{2b}=u^2+(u-1)^2$ because $(u,v)=1$ there are $k,l$ so as $(k,l)=1$ $5^b=k^2+l^2$ $u=2kl$ $u-1=k^2-l^2$ So we have to solve the system ... $k^2-l^2+1=2kl$ $k^2+l^2=5^b$ From first we have that $k=l+\sqrt{2l^2-1}$ So putting k at $k^2+l^2=5^b$ finally we will have the equation.. $5^{2b}+8l^4+1+2*5^b=8l^2*5^b+4l^2$ so to have solutions must $5^b<8l^2$ and because $5^b=k^2+l^2$ finally must $k^2<8l^2$ and from the equation $k^2-l^2+1=2kl$ devide with $l^2$ finally see that $2k<7$ so $k<4$ and only for $k=2$ , $l=1$ , $b=1$ $x=y=2$ we have solution.. ****EDITED typo $5^{2b-1}$ to $5^{2b}-1$

I have a little ugly solution with standard methods: $7^x- 2\cdot 5^y =-1 \implies (-1)^x \equiv 1 \pmod 4 \implies 2|x$.Also we obtain $(-1)^y \equiv 1 \pmod 3 \implies 2|y$.Now we try in cases $y=0$ and $y=2$ and we get solutions $(x,y)=(0,0)$ and $(x,y)=(2,2)$.Now we consider for $y \ge 4$.Now we put $x=2k$ and $y=2l$ and get $49^k + 2 \cdot 25^l=-1 \implies 49^k \equiv 124 \pmod{125}$.By calculating all the cases from $1-10$ we get that $k \equiv 5 \pmod{10}$ and we put $k=10x+5$.Now we have $(49^5)^{2x+1} - 2 \cdot 25^l=-1 \implies 49^5-1 | 2 \cdot 25^l \implies 49^4-49^3+49^2-49^1+1|5^{2(l-1)} $.Now by just easily calculating the left side we get a contradiction because it isn't power of $5$ so the only solutions are $\boxed {(x,y)=(0,0)}$ and $\boxed{(x,y)=(2,2)}$.

Notice that $x=0$ yields $1-2\cdot5^y=-1\Longrightarrow 5^y=1\Longrightarrow y=0$. So $(x,y)=(0,0)$ is a solution. So from now on $x,y\ge1$ We can rewrite the expression as $7^x+1=2\cdot5^2$ Furthermore, by checking $\pmod 3$ we obtain $7^x+1\equiv 2\pmod 3\text{ and } 2\cdot5^y\equiv 2(-1)^y\pmod 3$ thus $y$ must be even. Also notice that if we take $\pmod 4$ we obtain $7^x+1\equiv (-1)^x+1\pmod 4\text{ and } 2\cdot5^y\equiv 2\pmod 4$ thus $x$ is also even. Not let $x=2n\text{ and }y=2k$, thus the equation transforms into $49^n+1=2\cdot25^k$ Furthermore, by taking $\pmod {50}$ we obtain $2\cdot25^k\equiv 0\pmod {50}\text{ and } 49^n+1\equiv (-1)^n+1\pmod {50}$ thus $n$ is odd. Therefore we can rewrite the expression as $49^n-(-1)^n=2\cdot25^k$, and we can use $LTE$. $\nu_5\left(49^n-(-1)^n\right)=\nu_5(2\cdot5^{2k})\Longrightarrow \nu_5(n)+\nu_5(50)=2k\Longrightarrow \nu_5(n)=2(k-1)$ thus $n$ is divisible by $5$ if $k>1$. So $FTSOC$ assume $k>1$. Moreover let $n=10t+5$ since $n$ is odd. Now we can rewrite the equation as $(49^5)^{2t+1}+1=2\cdot25^k\Longrightarrow (49^5+1)((49^5)^{2k}-\dots+1)=2\cdot25^k\Longrightarrow 49^5+1|2\cdot25^k$ furthermore after some computation we obtain $141237625|5^{2k}\Longrightarrow 5^3\cdot281\cdot4021|5^{2k}\Longleftrightarrow \frac{5^{2k-3}}{281\cdot4021}\in\mathbb{Z}$ which is a contradiction since $281\cdot4021$ is not a power of $5$. Thus $k=1$ So the equation boils down to $49^n+1=50\Longrightarrow n=1$ Thus $n=1$ Furthermore we get that $(x,y)=(2,2)$ is the only other solution. Now, to sum up $\boxed{(x,y)=(0,0),(2,2)}$ are the only solutions in nonnegative integers $\blacksquare$.

So easy, Guys its LTE ,just profe $2$ divide $x$ but $4$ not, $49^k+1=2•5^y$ by lte $V_p(k)=y-2$

We can easily obtain by $$\text{mod3 ; mod4 ; mod5 ; mod7 ; mod11 ; mod13 ; mod19 ; mod41 ; mod125}$$