Disclaimer: this answer makes some coarse simplifications in order to keep it focussed on the crucial points. If you feel like pointing out something about orbit eccentricities or whatever in the comments, by all means go ahead, but I won't reply.
You can travel to Mars in a straight line, it's just not efficient.
Radially
This is perhaps what most people think how moving between different-height orbits should work: you fly away from the sun in a straight line. After all, our problem is that we're too low, too close to the sun, and we want to get further away, so surely the way to accomplish that is to fire our thrusters to give us a heading straight away from the sun? Then we'll move in a line until our kinetic energy is exchanged for potential energy. Easy.
Well, this would actually be the best way if we had started from a platform that was somehow moored at 1 AU from the sun, and tried to get to another platform moored at 1.5 AU. But clearly, no such platforms exist – anything just sitting there would quickly fall into the sun itself. (Short of being fixed to a Dyson sphere or something).
Instead, we start from Earth which is buzzing around the sun at 30 km⁄s, and before we could start heading out towards Mars radially we would need to brake from this velocity – but braking takes just as much fuel as accelerating to 30 km⁄s from a standstill. So that's crazy expensive – 30 km⁄s is a huge amount of $\Delta v$, and that's before we would even start our height-changing maneuver. And then once at Mars' orbit, we would need to accelerate again to match its velocity, before we could land there.
For the “lifting” itself, there are different ways you can do it. The most efficient way (like we cared about efficiency...) is to do it all in one quick burn at the start to give you enough kinetic energy so the radial velocity will zero again just when you reach Mars. Any other option will either waste time (and time=fuel=money if you're about to fall into the sun...), or require wasting even more $\Delta v$ on braking down the radial velocity at Mars.
Total cost: 30 km⁄s + √((42 km⁄s)2 - (34 km⁄s)2) + 24 km⁄s ≈ 79 km⁄s
That's absurdly expensive.
Fortunately, braking down from Earth's orbital velocity is not only unnecessary, it's actually completely counterproductive because that kinetic energy can be used to get to a higher orbit.
Tangentially
The next best idea you might have is to just accelerate outwards without braking from orbital velocity first. But that means the sun's gravity will not only slow down your outbound speed like it did in the radial case, but also bend your trajectory. For small orbital changes this is actually not noticeable. This is how for instance a spacecraft maneuvers right before docking to the ISS: you want to move the last meters closer, you simply fire the thusters in the opposite direction as one would intuitively expect, and that sends you towards the station. Never mind that the orbit bends a bit in the few seconds before you actually get there.
But Mars is kinda a bit more that a few meters away from Earth[citation needed], so in this case the bend would be noticeable. Now, you could theoretically prevent that by not firing in merely a short burn at the start, but instead steadily at always the same acceleration as the sun's gravity imposes on you, thus exactly cancelling the curving effect. As a result, you would move outbound in a straight line tangentially to Earth's orbit. The time-dependent distance to the sun would be
$$
R_\text{tngt}(t) = \sqrt{r_{\text{Earth}}^2 + (t\cdot v_\text{Earth})^2}
$$
Thus the travel time would be
$$
T_\text{tngt} = \frac{\sqrt{r_{\text{Mars}}^2 - r_{\text{Earth}}^2}}{v_\text{Earth}} \approx 5.77\cdot 10^6\:\mathrm{s}
$$
or 67 days. The fuel for that ride takes
$$\begin{align}
\Delta v_\text{tngt}
=& \int\limits_0^{T_\text{tngt}}\!\!\!\!\mathrm{d}t
\: \frac{G\cdot m_\text{Sun}}{(R_\text{tngt}(t))^2}
\\=& \frac{G\cdot m_\text{Sun}}{r_{\text{Earth}}^2}
\cdot \int\limits_0^{T_\text{tngt}}\!\!\!\!\mathrm{d}t
\: \frac1{1 + (t\cdot \frac{v_\text{Earth}}{r_{\text{Earth}}})^2}
\\=& \frac{G\cdot m_\text{Sun}}{v_{\text{Earth}}\cdot r_{\text{Earth}}}
\cdot \int\limits_0^{T_\text{tngt}\cdot \frac{v_\text{Earth}}{r_{\text{Earth}}}}\!\!\!\!\mathrm{d}\tau
\: \frac1{1 + \tau^2}
\\=& \frac{G\cdot m_\text{Sun}}{v_{\text{Earth}}\cdot r_{\text{Earth}}}
\cdot \arctan\left(T_\text{tngt}\cdot \frac{v_\text{Earth}}{r_{\text{Earth}}}\right)
\\=& \frac{G\cdot m_\text{Sun}}{v_{\text{Earth}}\cdot r_{\text{Earth}}}
\cdot \arctan\left(\sqrt{\frac{r_{\text{Mars}}^2}{r_{\text{Earth}}^2} - 1}\right)
\end{align}$$
which comes out to 25.5 km⁄s. That's not quite as insane as the radial approach, but still already much worse than a Hohmann transfer, and we're not finished yet: we still need to match Mars' orbital velocity, which involves getting rid of our radial velocity component and adding prograde velocity.
The latter bit indicates what we're doing wrong all the time: letting the sun bend our trajectory is actually a good thing, and what we want is prograde acceleration. The consequence is exactly what you do in a Hohmann transfer: don't bother with radial maneuvers at all, but only use two prograde burns and a freefall elliptical trajectory in between.
