With a few reasonable assumptions, an answer can be found with a minimum of brute force work. Note first that for each square, cube, and fourth power to be found, if said number divides one of the consecutive numbers, then each prime factor of that number also divides that particular number. Note also that three consecutive numbers, which I will represent as $y-1,y,y+1$, will have one member divisible by $3$, and one or two that are even.
I will make the simplifying assumption that among the numbers that are solutions, one of them will have $3$ as a prime factor, and another will have $2$ as a prime factor. This is not necessarily the only possible path to solutions, but it seems a reasonable path to solutions of modest magnitude. In this case, the product of the three consecutive numbers, $y^3-y$, will be divisible by $72$, since one of the numbers will have a factor of $3^k$ where $k$ is $2,3,4$, and there will have to be at least $3$ factors of $2$ among the numbers (at least $2^3$ if $2\mid y$ and at least $2^3$ if $2\mid (y+1)(y-1)$). Finally, there will be at least one more prime $p$ such that $p^2\mid (y^3-y)$. I will make the further simplifying assumption that $p^2\mid (y-1)$ so as to avoid the necessity to look at $p^3\mid y$ or $p^4\mid (y+1)$ for large values of $p$.
Using a simple spreadsheet, I looked at $\frac{y^3-y}{72p^2}$ for $1<y\le 500$ and $p=5,7,11,13$ to identify candidates for $y$. Choosing only results that were integers, I further culled the results for those where $p^2\mid (y-1)$.
For $p=13$ no candidates were found. For $p=11$, candidates $243,485$ were identified, but ruled out by examination with regard to the divisibility requirements. For $p=7$, candidates $99,197$ were identified, but ruled out by examination with regard to the divisibility requirements. For $p=5$, candidates $251,351,451$ were identified. Of these, only $351$ met the divisibility requirements.
So within the constraints of my assumptions, up to $500$, the only solution is $(y-1),y,(y+1)=350,351,352$.
Best Answer
Hint: It's a good idea to think about which primes these primes could be. For example, since you're looking for small numbers, you probably want to see what happens if you stipulate that the prime $p$ so $p^4$ divides your largest number is $2$ -- then try for $3$, $5$, etc. until you find that your result ends up being too big. I'll show you a worked example for the simpler problem of finding two consecutive numbers the first of which has a perfect square of a prime as a factor and the second of which has the perfect cube of a prime as a factor:
Let our integers be $a$ and $a+1$. We know that either $8|a+1$ (so $2$ is our prime) or $a+1\geq 27$. Let's first try the case $8|a+1$. Then we want to find a number that the perfect square of a prime divides in the sequence
$$7, 15, 23, 31, 39, 47, 55, 63, 71...$$
and we find that the first one occurs at $63$, so ($63,64$) is a solution. Now let's try when $27|a+1$. We get that $a$ is in
$$26, 53, 80...$$
and then everything above $53$ is too big, since we've already found the example $(63,64)$. Now, if our prime is any bigger than $3$, we get that $a+1\geq 125$, which is going to give us something much larger than $(63,64)$, and thus $(63,64)$ is our answer.