Solve $3^a-5^b=2$ for integers a and b.

So I have got that (a,b)=(1,0),(3,2) are solutions for the eqations, and maybe the only one.

We know that $b$ is even (since $2^b+2$ is divisible by 3). We also know that the only solution to $y^2+2=x^3$ is $y=5,x=3$. (Solving the diophantine equation $y^{2}=x^{3}-2$)

Thus it is sufficient to show that $a$ is divisible by 3. Suppose that $a \geq 2$. Since 9 divides $5^b+2$, we get that $b=6k+2=3m+2$. We have $25(125)^m+2=3^a$. We get $3^a$ is $27$ mod $31$ which forces $a$ to be $3$ mod $30$, in particular divisible by $3$.

These are the only solutions modulo $$ 5^3 \cdot 601. $$ If you had a solution with $b \geq 3$ then we would require that $a \equiv 43$ modulo $100$. Modulo $601$ (there are lots of other choices), there are only $12$ choices for $5^b$. Solving $3^a \equiv 5^b+2 \mod{601}$ for each of these choices tells us that necessarily $a$ is congruent to $0, 1$ or $3$ modulo $75$, a contradiction.