What does unique factorization exactly mean here? Given $a$ and $b$ are integers:
$$ 3\cdot 2 \cdot 5^2 \cdot 7^a = 2 \cdot 3^2 \cdot 7 \cdot 5^b $$
The answer is as follows:

Since both sides of this equation are integers and have unique factorizations, it follows that $a = 1$ and $b = 2$ is the only solution.

What does unique factorization exactly mean here? I looked up the word unique factorization on google but I don't really understand it. Can you explain it at a pre-calculus level? Also, what aspect of 'integers' makes the solution $a = 1$, and $b = 2$ ??

Every positive integer is either prime or composite or the number one.
If the number is composite then it can be factored as a product of smaller numbers.  Ex:  $120 = 12\cdot10$ or we could say $120 = 30\cdot4$ or we could say $120 = 6\cdot10\cdot2$.
If we factor a number into smaller composites and then factored those number we'd eventually factor it do to a factorization containing only primes.  Ex:  $120 = 12\cdot10 = (3\cdot4)\cdot(2\cdot5) = 3\cdot2\cdot2\cdot2\cdot5$  or we could say $120 = 30\cdot4 =(6\cdot5)\cdot(2\cdot2) = 2\cdot3\cdot5\cdot2\cdot2$ or $120=6\cdot10\cdot2 = 2\cdot3\cdot2\cdot5\cdot2$.
And if we have multiple occurrences of the same prime factor we can list them as a power.  Ex: $120 = 3\cdot2^3\cdot5$ or $120 = 2^3\cdot3\cdot5$ or $120 =3\cdot5\cdot2^3$.
Now perhaps the most anti-climatic conclusion you'll ever see.
No matter how you break a number down into primes you will always and up with the same primes and the same number of each of them.  Each number has only one unique way of being broken down into primes.
And that is what unique prime factorization means.
So if you have $3^2 \cdot 2 \cdot 5^2 \cdot 7^a = 2 \cdot 3^2 \cdot 7 \cdot 5^b$ then you have a number.  By the Left-Hand side we know it factors into: two $3$s, one $2$, two $5$s, and some $7$ but we don't know how many.  On the RHS we see that it factors down into: one $2$, two $3$s, one $7$, and some $5$s but we don't know how many.
But by the Unique prime factorizations we know the two lists of components must be exactly the same.  So how many $7$s are there?  Well, by the RHS we know there is one $7$. And how many $5$s are there?  Well, by the LHS we know there are two $5$s.
So that's that.  If $3^2 \cdot 2 \cdot 5^2 \cdot 7^a = 2 \cdot 3^2 \cdot 7 \cdot 5^b$ then $a =1$ and $b = 2$ and then number is $3^2 \cdot2\cdot5^2\cdot7 = 9\cdot2\cdot5^2\cdot7 = 3150$.

Note: The really hard way to do it:
$3^2 \cdot 2 \cdot 5^2 \cdot 7^a = 2 \cdot 3^2 \cdot 7 \cdot 5^b$  Divide both sides by $3^2\cdot2$
$5^2\cdot7^a = 7\cdot5^b$.  Divide both sides by $7$.
$5^2\cdot7^{a-1} = 5^b$.  Divide both sides by $5^2$.
$7^{a-1} = 5^{b-2}$.  
So.... the question is: what (integer) power can you raise $5$ to get a (integer) power of $7$ (or vice versa).  The answer is: $5$ and $7$ are prime; there isn't any such power. The only way this makes sense is if  $7^0  = 5^0 = 1$.  So $a-1 = b-2 =0$.
The point being: if you have ever told that  $3^a\cdot5^b = 3^c\cdot5^d$ and you are told $a$ and $b$ are  integers, then it should be immediately clear $a = c$ and $b = d$.
