Here is an approach that's reasonably effective without needing to use the concept of logarithms, or requiring clever insights that would likely confuse the sort of audience that doesn't understand logs or have a calculator that computes them.
We begin with the familiar observation that 2^3 is just a little smaller than 3^2. Compute 8^k/9^k (by repeated multiplication since we've been told our calculator doesn't know how to compute powers) for smallish k until we find that 2^18/3^12 is about 0.493, just slightly less than 1/2. So 2^19/3^12 ~= 0.986. It's actually nearer 0.9865.
(Remark: this is why there are 12 semitones in an octave: the "circle of fifths", with ideal frequency ratios 3:2, almost closes up after 12 steps.)
Now 2017 is a little more than 106x19. So the next thing we do is to say that 2^2014 / 3^1272 ~= 0.9865^106.
We could compute this last thing without too much pain even on a nasty 4-function calculator by repeated squaring and find that it's somewhere around 0.23. More precisely: First square once: we want 0.9732^53. That's about 3% more than 0.9732^54. 54 is a nice number. Cube, cube, cube, square. (Our calculator probably doesn't have a button for that, but anyone can remember four digits for long enough to do it by multiplying.) That gives us about 3% less than 0.23. It will soon become apparent that neither that 3% nor the other accumulated rounding errors are big enough to matter.
So then we know that 2^2017 / 3^1274 ~= 8/9 x 0.23, so 2^2017 = 3^1274 x 0.2ish, so 2^2017 is comfortably between 3^1272 and 3^1273.
So, how many operations do we need to do on our nasty little 4-function calculator? Well, it depends on how much we can do in our head. Something like this:
- Enter 8/9 * 8/9 = * 8/9 = ... (counting by hand) until we see something very close to a nice number, which in this case is 1/2. We'll end up with six lots of 8/9, so 28 keystrokes.
- Enter * 2 = to get our estimate of 2^19/3^12. 31 keystrokes so far. Remember the number 0.9865 mentally.
- Divide 2017 by 19 to get 106-and-a-bit: 8 keystrokes. 39 keystrokes so far.
- Now let's figure out 0.9865^106 using the procedure above. I'll assume we don't have a "square" or "cube" button; I'll work to four figures, or 3 when the error is obviously going to be small; and I'll assume I have enough brain to be able to square e.g. 9865 instead of 0.9865 and figure out what the result means. So I do 9865 * 9865 = (10 keystrokes); then 973 * 973 * 973 = (12 keystrokes); then 921 * 921 * 921 = (12 keystrokes); then 781 * 781 * 781 = (12 keystrokes); then 476 * 476 = (8 keystrokes). That's 54 keystrokes, so 93 keystrokes so far.
- Now what we know is that 2^2017 = 3^1274 * 0.22ish * 8/9. We can see without needing any actual calculation that 0.22 * 8/9 is bigger than 1/9 but smaller than 1/3, and that there's way more slack in there than could be eroded by rounding errors, so we know the answer we need is 1272.
Some 4-function calculators behave in ways that would substantially reduce the number of keystrokes needed. For instance, on some I think entering
*== computes the cube of a number. For these we could do the work of those last 54 keystrokes faster and more accurately by typing
.9865*=*==*==*==*= for 18 keystrokes, saving 36 keystrokes. Similarly, the first step would take not 28 keystrokes but 9 keystrokes. I assume we are still too stupid to work out 2017/19 mentally, though, so we still need that calculation and it still means that we no longer have .9865... available without rekeying it. Still, this ends up taking 9+3+8+18=38 keystrokes instead of 93.
OP is keen to know how many times I have to press "=" with this approach. The number appears to be 5 when finding that 2^18/3^12 is a good waypoint, plus 1 when doubling, plus 1 when finding out how many 19s go into 2017, plus 5 when taking a 108th (~= 106th) power. Total is 12. I'm not entirely convinced that the number of "="s is a great metric; for instance, I could reduce that last 5 to 1 by doing a whole lot more multiplication operations, but that would not in any useful sense be an improvement.
OP also wishes to know how many operator keys I have to press. I'm not sure whether he wants the = key counted there; I will include it. Then there are 10+2+2+13=27 operator key presses, if I have done my arithmetic correctly.